Free Math Worksheets

- What does “right triangle” tell you about \(\gamma\)? - What is the sum of the interior angles of a triangle? - Which angle measures are already known?

1. Since the triangle is a right triangle, \(\gamma = 90^\circ\). 2. The interior angles of a triangle total \(180^\circ\). 3. Therefore, \(\beta = 180^\circ - 90^\circ - 37.5^\circ = 52.5^\circ\).

\(\beta = 52.5^\circ\) and \(\gamma = 90^\circ\).

- What is the sum of the interior angles of a triangle? - Add the two known angles. - Subtract their sum from \(180^\circ\).

1. The interior angles of a triangle total \(180^\circ\). 2. Write \(48^\circ + 74^\circ + \beta = 180^\circ\). 3. The known angles total \(122^\circ\), so \(\beta = 180^\circ - 122^\circ = 58^\circ\).

\(\beta = 58^\circ\)

- What does a right angle measure? - What is the sum of the two acute angles in a right triangle? - Use the triangle angle sum.

1. The angle at \(C\) measures \(90^\circ\). 2. The interior angles of a triangle total \(180^\circ\). 3. Therefore, \(\alpha = 180^\circ - 90^\circ - 62^\circ = 28^\circ\).

\(\alpha = 28^\circ\)

- You know the final measure and the amount of increase. How can you reverse the change? - Write an equation in the form original angle plus change equals final angle.

1. Write the equation \(\beta + 18^\circ = 112^\circ\). 2. Subtract \(18^\circ\) from both sides: \(\beta = 112^\circ - 18^\circ = 94^\circ\).

\(\beta = 94^\circ\)

- Adjacent angles can combine to form a larger angle. - Which operation combines two parts to find a whole?

1. Because the angles are adjacent, their measures add to the measure of the entire angle. 2. Calculate: \(\gamma = \alpha + \beta = 25^\circ + 45^\circ = 70^\circ\).

The measure of \(\gamma\) is \(70^\circ\).

- How many congruent angles make up the larger angle? - Which operation finds one equal share of a total?

1. The \(114^\circ\) angle is divided into three angles of equal measure. 2. Divide the total measure by \(3\): \(114^\circ \div 3 = 38^\circ\).

Each smaller angle measures \(38^\circ\).

- What is the sum of the three interior angles of a triangle? - Subtract the two known angle measures from that total.

1. The interior angles of a triangle sum to \(180^\circ\). 2. Therefore, \(\gamma = 180^\circ - 55^\circ - 65^\circ = 60^\circ\).

\(\gamma = 60^\circ\)

- In which smaller triangle is the unknown angle located? - Which two angle measures are known in that triangle?

1. Consider triangle \(APC\). 2. Its two known angles measure \(30^\circ\) and \(20^\circ\). 3. Use the triangle angle sum: \(\angle APC = 180^\circ - 30^\circ - 20^\circ = 130^\circ\).

\(\angle APC = 130^\circ\)

- What is the total measure of the three interior angles of a triangle?

1. The interior angles of a triangle sum to \(180^\circ\). 2. Therefore, \(\gamma = 180^\circ - 33^\circ - 112^\circ = 35^\circ\).

\(\gamma = 35^\circ\)

- What does the right-angle mark at \(C\) tell you?

1. The two acute angles of a right triangle sum to \(90^\circ\). 2. Therefore, \(\beta = 90^\circ - 25^\circ = 65^\circ\).

\(\beta = 65^\circ\)

- What is the sum of the three interior angles of a triangle? - How many degrees remain after subtracting the known angle? - What measurements define acute, right, and obtuse angles?

1. The interior angles of a triangle have a sum of \(180^\circ\). 2. Since one angle measures \(110^\circ\), the other two angles have a sum of \(180^\circ - 110^\circ = 70^\circ\). 3. Each angle in a triangle must be greater than \(0^\circ\), so each of the two remaining angles must be less than \(70^\circ\). 4. Any angle less than \(90^\circ\) is acute. Therefore, both remaining angles must be acute.

Both remaining angles must be acute. Together they measure \(70^\circ\), and each one must be greater than \(0^\circ\), so each is less than \(70^\circ\) and therefore acute.

- What is the sum of the interior angles of a quadrilateral? - Add the given angle measures in each part. - Compare the total in part b with the required quadrilateral angle sum.

1. The interior angles of any quadrilateral total \(360^\circ\). 2. For part a, \(\delta = 360^\circ - (85^\circ + 110^\circ + 45^\circ) = 360^\circ - 240^\circ = 120^\circ\). 3. For part b, the claimed angles total \(100^\circ + 120^\circ + 80^\circ + 70^\circ = 370^\circ\). 4. Since \(370^\circ \ne 360^\circ\), those angles cannot form a quadrilateral.

a) \(\delta = 120^\circ\) b) No. The four angles total \(370^\circ\), not \(360^\circ\).

- What is the sum of the interior angles of a triangle? - What is true about the base angles of an isosceles triangle? - When is a triangle classified as obtuse? - Pay attention to whether the given angle is a vertex angle or a base angle.

1. In the first triangle, the two base angles total \(180^\circ - 40^\circ = 140^\circ\). 2. The base angles of an isosceles triangle are congruent, so each measures \(140^\circ \div 2 = 70^\circ\). The angles are \(40^\circ\), \(70^\circ\), and \(70^\circ\). 3. In the second triangle, both base angles measure \(40^\circ\). The vertex angle is \(180^\circ - (40^\circ + 40^\circ) = 100^\circ\). 4. The second triangle is obtuse because it has an angle greater than \(90^\circ\).

The first triangle has missing angles of \(70^\circ\) and \(70^\circ\). The second triangle has missing angles of \(40^\circ\) and \(100^\circ\). The second triangle is obtuse.

- What is the sum of the interior angles of a triangle? - How can angle measures show that a triangle is isosceles? - What angle measure makes a triangle a right triangle?

1. For part a, \(\gamma = 180^\circ - (42^\circ + 96^\circ) = 42^\circ\). Since \(\alpha = \gamma\), the triangle is isosceles. 2. For part b, \(\alpha = 180^\circ - (25^\circ + 130^\circ) = 25^\circ\). Since \(\alpha = \beta\), the triangle is isosceles. 3. For part c, \(\beta = 180^\circ - (45^\circ + 90^\circ) = 45^\circ\). Since \(\alpha = \beta\), the triangle is isosceles. Since \(\gamma = 90^\circ\), it is also a right triangle.

a) \(\gamma = 42^\circ\); isosceles b) \(\alpha = 25^\circ\); isosceles c) \(\beta = 45^\circ\); isosceles and right

- What is the sum of the interior angles of every triangle? - What is true about the base angles of an isosceles triangle? - What are the angle measures in an equilateral triangle? - A quick sketch may help you test each statement.

1. For a), two right angles have a sum of \(90^\circ + 90^\circ = 180^\circ\). That would leave \(0^\circ\) for the third angle, so the statement is false. 2. For b), if the vertex angle is \(120^\circ\), the two congruent base angles have a sum of \(180^\circ - 120^\circ = 60^\circ\). Each base angle is \(60^\circ \div 2 = 30^\circ\), so such a triangle exists. The statement is true. 3. For c), the three angles of an equilateral triangle are congruent. Each angle measures \(180^\circ \div 3 = 60^\circ\). Since all three angles are less than \(90^\circ\), the triangle is acute. The statement is true.

a) False b) True c) True

- What is the total turn after one complete trip around a convex polygon? - How are an exterior angle and its adjacent interior angle related? - Subtract the known turns from the full turn.

1. The exterior turns of any convex polygon total \(360^\circ\). 2. The fourth exterior angle is \(360^\circ - (85^\circ + 105^\circ + 90^\circ) = 80^\circ\). 3. An interior angle and its adjacent exterior angle are supplementary. 4. Therefore, the fourth interior angle is \(180^\circ - 80^\circ = 100^\circ\).

a) \(80^\circ\) b) \(100^\circ\)

- Subtract the given angle from \(360^\circ\). - A reflex angle is greater than \(180^\circ\) and less than \(360^\circ\).

1. A full circle measures \(360^\circ\). 2. The remaining angle is \(\alpha=360^\circ-75^\circ=285^\circ\). 3. Because \(180^\circ<285^\circ<360^\circ\), \(\alpha\) is a reflex angle.

\(\alpha=285^\circ\), a reflex angle

- A straight angle measures \(180^\circ\). - Add the measured angle to the straight angle.

1. A straight angle measures \(180^\circ\). 2. The reflex angle is the straight angle plus the measured acute angle: \(180^\circ+42^\circ=222^\circ\).

\(222^\circ\)

- Compare each obtuse angle with \(90^\circ\). - Decide how the total of four angles greater than \(90^\circ\) compares with \(360^\circ\).

1. If all four angles were obtuse, each would be greater than \(90^\circ\). 2. Their sum would therefore be greater than \(4 \cdot 90^\circ=360^\circ\). 3. This contradicts the quadrilateral angle sum, so the claim is false.

No. Four obtuse angles would have a sum greater than \(360^\circ\).

- The interior angles sum to \(360^\circ\). - Subtract the three known angles from the total. - Compare the result with \(90^\circ\).

1. The three known angles have a sum of \(90^\circ+90^\circ+90^\circ=270^\circ\). 2. The fourth angle is \(360^\circ-270^\circ=90^\circ\). 3. The fourth angle is a right angle, so all four angles are right angles. The quadrilateral cannot have an obtuse angle.

a) \(90^\circ\) b) Right angle c) No. All four angles must be right angles.

- What is the defining property of an angle bisector? - How are the two smaller angles related? - Add the two parts to find the full angle.

1. An angle bisector divides an angle into two congruent angles. 2. Therefore, the angle between \(w\) and \(h\) is equal to the given \(28^{\circ}\) angle between \(g\) and \(w\), so \(\alpha = 28^{\circ}\). 3. The full angle is \(\beta = 28^{\circ} + 28^{\circ} = 56^{\circ}\).

\(\alpha = 28^{\circ}\) and \(\beta = 56^{\circ}\)

- A full turn measures \(360^\circ\). - Subtract the interior angle from the full turn. - A reflex angle is between \(180^\circ\) and \(360^\circ\).

1. The interior angle and the reflex angle make a full turn of \(360^\circ\). 2. Therefore, \(\alpha=360^\circ-135^\circ=225^\circ\). 3. Since \(180^\circ<225^\circ<360^\circ\), \(\alpha\) is a reflex angle.

1. Reflex angle 2. \(\alpha=225^\circ\)

- What is the interior angle sum of a quadrilateral? - Add the three known angles. - Subtract that total from \(360^\circ\).

1. The interior angles of any quadrilateral total \(360^\circ\). 2. Write \(75^\circ + 110^\circ + 85^\circ + \delta = 360^\circ\). 3. The three known angles total \(270^\circ\). 4. Therefore, \(\delta = 360^\circ - 270^\circ = 90^\circ\).

\(\delta = 90^\circ\), by the quadrilateral interior angle sum theorem.

- What is the interior angle sum of a quadrilateral? - Add the three known angles. - The parallel bases also make the two angles on the right leg supplementary.

1. The interior angles of any quadrilateral total \(360^\circ\). 2. The three known angles total \(90^\circ + 90^\circ + 115^\circ = 295^\circ\). 3. Therefore, \(\alpha = 360^\circ - 295^\circ = 65^\circ\).

\(\alpha = 65^\circ\)

- What does an angle bisector tell you about the two smaller angles? - Find the full angle between the rightward ray of \(g\) and \(s\). - What sum do angles in a linear pair have?

1. Since \(w\) is an angle bisector, the two smaller angles are congruent. Therefore, \(\alpha=55^\circ\). 2. The full angle between the rightward ray of \(g\) and \(s\) is \(55^\circ+55^\circ=110^\circ\). 3. This \(110^\circ\) angle and \(\beta\) form a linear pair, so \(\beta=180^\circ-110^\circ=70^\circ\).

a) \(\alpha=55^\circ\). b) \(\beta=70^\circ\).

- What is true about the base angles of an isosceles triangle? - Add the two known base angles. - How much is needed to reach \(180^\circ\)?

1. The two base angles measure \(71^\circ\) each. 2. The interior angles of a triangle total \(180^\circ\). 3. Therefore, \(\gamma = 180^\circ - 71^\circ - 71^\circ = 38^\circ\).

\(\gamma = 38^\circ\)

- How are the three smaller angles related to the entire angle? - Subtract the measures of the two known parts from the total.

1. The three angle measures add to \(125^\circ\): \(\alpha + \beta + \gamma = 125^\circ\). 2. Substitute the known measures: \(32^\circ + 48^\circ + \gamma = 125^\circ\). 3. Add the known angles: \(32^\circ + 48^\circ = 80^\circ\). 4. Subtract: \(\gamma = 125^\circ - 80^\circ = 45^\circ\).

\(\gamma = 45^\circ\)

- Identify which given angles contain or combine to form each unknown angle. - Decide whether each unknown angle requires addition or subtraction.

1. Ray \(b\) lies inside the \(75^\circ\) angle between rays \(a\) and \(c\). Therefore, the angle between rays \(a\) and \(b\) is \(75^\circ - 25^\circ = 50^\circ\). 2. The angle between rays \(b\) and \(d\) is the sum of the \(25^\circ\) angle between rays \(b\) and \(c\) and the \(35^\circ\) angle between rays \(c\) and \(d\). Therefore, it measures \(25^\circ + 35^\circ = 60^\circ\).

The angle between rays \(a\) and \(b\) is \(50^\circ\), and the angle between rays \(b\) and \(d\) is \(60^\circ\).

- A straight angle measures \(180^\circ\). - All three adjacent angle measures must add to \(180^\circ\).

1. The three adjacent angles form a straight angle, so their measures sum to \(180^\circ\). 2. Subtract the two known measures: \(180^\circ - (55^\circ + 65^\circ) = 180^\circ - 120^\circ = 60^\circ\).

The middle angle measures \(60^\circ\).

- A straight angle measures \(180^\circ\). - Represent the two equal outer angles with the same variable.

1. The three adjacent angles form a straight angle: \(\alpha + 90^\circ + \gamma = 180^\circ\). 2. Since \(\alpha = \gamma\), write \(2\alpha + 90^\circ = 180^\circ\). 3. Then \(2\alpha = 90^\circ\), so \(\alpha = 45^\circ\).

\(\alpha = 45^\circ\)

- Into how many triangles can a hexagon be divided from one vertex? - In a regular polygon, the interior-angle sum is divided equally among all vertices.

1. The sum of the interior angles of a hexagon is \((6 - 2) \cdot 180^\circ = 720^\circ\). 2. A regular hexagon has six congruent interior angles, so \(\alpha = 720^\circ \div 6 = 120^\circ\).

\(\alpha = 120^\circ\)

- What is the measure of each interior angle of a square? - What is the measure of each interior angle of an equilateral triangle? - How do the two angles at vertex \(B\) combine?

1. Each interior angle of a square measures \(90^\circ\), so \(\angle ABC = 90^\circ\). 2. Each interior angle of an equilateral triangle measures \(60^\circ\), so \(\angle ABE = 60^\circ\). 3. Therefore, \(\alpha = \angle ABC - \angle ABE = 90^\circ - 60^\circ = 30^\circ\).

\(\alpha = 30^\circ\)

- Subtract the vertex angle from \(180^\circ\). - The two base angles are congruent, so divide the remaining measure equally.

1. The base angles of an isosceles triangle are congruent, so \(\beta = \gamma\). 2. Subtract the vertex angle from the triangle angle sum: \(180^\circ - 40^\circ = 140^\circ\). 3. Divide the remaining measure equally: \(\beta = \gamma = 140^\circ \div 2 = 70^\circ\).

\(\beta = 70^\circ\) and \(\gamma = 70^\circ\)

- All adjacent angles on one side of a straight line form a \(180^\circ\) angle.

1. The adjacent angles on one side of a straight line sum to \(180^\circ\). 2. Write \(35^\circ + 55^\circ + \gamma = 180^\circ\). 3. Therefore, \(\gamma = 180^\circ - 90^\circ = 90^\circ\).

\(\gamma = 90^\circ\)

- How are an interior angle and its adjacent exterior angle related? - What does a pair of congruent angles imply about the opposite sides of a triangle?

1. The interior angle at \(C\) and its exterior angle form a linear pair, so \(\angle C = 180^\circ - 115^\circ = 65^\circ\). 2. Since \(\angle A = \angle C = 65^\circ\), the sides opposite those angles are congruent. Therefore, triangle \(ABC\) is isosceles. 3. The congruent legs are \(AB\) and \(BC\), and the base is \(AC\).

Triangle \(ABC\) is isosceles because \(\angle A = \angle C = 65^\circ\). Its base is \(AC\), and its legs are \(AB\) and \(BC\).

- First identify the base and the two congruent base angles.

1. The base angles are congruent, so the other base angle also measures \(54^\circ\). 2. Use the triangle angle sum: \(\gamma = 180^\circ - 54^\circ - 54^\circ = 72^\circ\).

\(\gamma = 72^\circ\)

- Divide the angle measure remaining after the vertex angle equally between the two base angles.

1. The two base angles are congruent. 2. Their total measure is \(180^\circ - 104^\circ = 76^\circ\). 3. Therefore, \(\alpha = \beta = 76^\circ \div 2 = 38^\circ\).

\(\alpha = 38^\circ\) and \(\beta = 38^\circ\)

- Which smaller angles combine to form \(\angle AOC\)? - Use subtraction to find each missing part. - Compare the two results.

1. Since \(\angle AOC = \angle AOB + \angle BOC\), \(\angle AOB = 80^\circ - 35^\circ = 45^\circ\). 2. Since \(\angle BOD = \angle BOC + \angle COD\), \(\angle COD = 80^\circ - 35^\circ = 45^\circ\). 3. The two unknown angles are congruent.

\(\angle AOB = 45^\circ\) and \(\angle COD = 45^\circ\). The angles are congruent.

- What is the measure of a right angle? - How can you represent the larger acute angle in terms of the smaller one? - What is the sum of the two acute angles in a right triangle?

1. A right angle measures \(90^\circ\). 2. The two acute angles must total \(180^\circ - 90^\circ = 90^\circ\). 3. Let \(x\) be the measure of the smaller acute angle. Then the larger acute angle measures \(3x\). 4. Write and solve the equation \(x + 3x = 90^\circ\): \(4x = 90^\circ\), so \(x = 22.5^\circ\). 5. The larger acute angle is \(3 \cdot 22.5^\circ = 67.5^\circ\).

The three angle measures are \(22.5^\circ\), \(67.5^\circ\), and \(90^\circ\).

- Express all three angles in terms of one variable, such as \(\alpha\). - How can you write “\(\alpha\) is \(15^\circ\) less than \(\beta\)” as an equation? - Use the sum of the interior angles of a triangle.

1. Let \(\alpha = x\). Then \(\beta = x + 15^\circ\) and \(\gamma = 2x\). 2. Use the triangle angle sum: \(x + (x + 15^\circ) + 2x = 180^\circ\). 3. Combine like terms and solve: \(4x + 15^\circ = 180^\circ\), so \(4x = 165^\circ\) and \(x = 41.25^\circ\). 4. Therefore, \(\alpha = 41.25^\circ\), \(\beta = 41.25^\circ + 15^\circ = 56.25^\circ\), and \(\gamma = 2 \cdot 41.25^\circ = 82.5^\circ\).

\(\alpha = 41.25^\circ\), \(\beta = 56.25^\circ\), and \(\gamma = 82.5^\circ\).

- What is the sum of the interior angles of a quadrilateral? - After subtracting the two known angles, how much remains for the two equal angles? - What relationship do opposite angles have in a parallelogram?

1. The two given angles total \(72.5^\circ + 107.5^\circ = 180^\circ\). 2. The interior angles of a quadrilateral total \(360^\circ\), so \(\gamma + \delta = 360^\circ - 180^\circ = 180^\circ\). 3. Since \(\gamma = \delta\), each angle measures \(180^\circ \div 2 = 90^\circ\). 4. In a parallelogram, opposite angles are congruent. Here, \(\alpha \ne \gamma\) and \(\beta \ne \delta\), so the quadrilateral cannot be a parallelogram.

a) \(\gamma = 90^\circ\) and \(\delta = 90^\circ\) b) No. Its opposite angles are not congruent.

- Draw diagonals from one vertex so that the pentagon is divided into nonoverlapping triangles. - How many triangles are formed? - Subtract the sum of the four known angles from the pentagon’s total angle sum.

1. Drawing diagonals from one vertex divides a pentagon into \(5 - 2 = 3\) triangles. 2. Therefore, the pentagon’s interior angle sum is \(3 \cdot 180^\circ = 540^\circ\). 3. The four known angles total \(112^\circ + 98^\circ + 125^\circ + 105^\circ = 440^\circ\). 4. The fifth angle is \(540^\circ - 440^\circ = 100^\circ\).

a) \(540^\circ\) b) \(100^\circ\)

- Express each unknown angle in terms of one variable. - Translate “twice” and “three times” into algebraic expressions. - What is the measure of a right angle? - Use the interior angle sum of a quadrilateral.

1. The interior angles of a quadrilateral total \(360^\circ\), so \(\alpha + \beta + \gamma + \delta = 360^\circ\). 2. Substitute the given relationships: \(\alpha + 2\alpha + 3\alpha + 90^\circ = 360^\circ\). 3. Combine like terms: \(6\alpha + 90^\circ = 360^\circ\). 4. Solve: \(6\alpha = 270^\circ\), so \(\alpha = 45^\circ\). 5. Then \(\beta = 2 \cdot 45^\circ = 90^\circ\) and \(\gamma = 3 \cdot 45^\circ = 135^\circ\).

\(\alpha = 45^\circ\), \(\beta = 90^\circ\), and \(\gamma = 135^\circ\).

- Consider a triangle with a right angle, then a triangle with an obtuse angle. - What happens to the angle sum if two angles are each at least \(90^\circ\)? - How much angle measure would remain for the third angle?

1. An acute triangle has three acute angles, so it has at least two. 2. A right triangle has one \(90^\circ\) angle. The other two angles have a sum of \(180^\circ - 90^\circ = 90^\circ\). Since both are greater than \(0^\circ\), each is less than \(90^\circ\), so both are acute. 3. An obtuse triangle has one angle greater than \(90^\circ\). The other two angles have a sum less than \(90^\circ\), so both are acute. 4. If a triangle had only one acute angle, the other two angles would each be at least \(90^\circ\). Their sum would already be at least \(180^\circ\), leaving no positive measure for the third angle. That is impossible.

The statement is true. An acute triangle has three acute angles, while a right triangle and an obtuse triangle each have exactly two acute angles. A triangle cannot have fewer than two acute angles because two nonacute angles would have a sum of at least \(180^\circ\).

- What is true about the base angles of an isosceles triangle? - What would happen if both base angles were obtuse? - At which vertex could an obtuse angle occur without violating the angle sum?

1. In an isosceles triangle, the two base angles are congruent. 2. Suppose a base angle were obtuse, so it measured more than \(90^\circ\). The other base angle would also measure more than \(90^\circ\). 3. Those two angles alone would have a sum greater than \(180^\circ\), which is impossible because all three interior angles of a triangle have a sum of exactly \(180^\circ\). 4. Therefore, an obtuse angle in an isosceles triangle can only be the vertex angle between the congruent sides. For example, a triangle with angle measures \(120^\circ\), \(30^\circ\), and \(30^\circ\) is isosceles and obtuse.

a) Yes, an isosceles triangle can have an obtuse angle. b) No. An obtuse base angle would force both congruent base angles to be greater than \(90^\circ\), so their sum would exceed \(180^\circ\). The obtuse angle must be the vertex angle.

- Imagine drawing a segment from one vertex to the opposite vertex. What shapes are formed? - Once you know the total angle measure, how can you split it into four equal parts? - Which quadrilaterals have four right angles?

1. A triangle has an interior angle sum of \(180^\circ\). A diagonal divides a quadrilateral into two triangles, so the interior angle sum of the quadrilateral is \(2 \cdot 180^\circ = 360^\circ\). 2. If all four angles have the same measure, divide the total by \(4\): \(360^\circ \div 4 = 90^\circ\). 3. A rectangle and a square each have four right angles.

a) The interior angle sum is \(2 \cdot 180^\circ = 360^\circ\). b) Each angle measures \(90^\circ\). Two possible quadrilaterals are a rectangle and a square.

- Represent congruent angles with the same variable. - For part b, use the smallest angle as the variable. - In each part, use the \(360^\circ\) interior angle sum of a quadrilateral.

1. For part a, use the quadrilateral angle sum: \(112^\circ + 48^\circ + \beta + \delta = 360^\circ\). 2. Since \(\beta = \delta\), \(160^\circ + 2\beta = 360^\circ\). Thus \(2\beta = 200^\circ\), so \(\beta = 100^\circ\) and \(\delta = 100^\circ\). 3. For part b, write \(\alpha + (\alpha + 10^\circ) + (\alpha + 20^\circ) + (\alpha + 30^\circ) = 360^\circ\). 4. Combine like terms: \(4\alpha + 60^\circ = 360^\circ\). Then \(4\alpha = 300^\circ\), so \(\alpha = 75^\circ\). 5. Therefore, \(\beta = 85^\circ\), \(\gamma = 95^\circ\), and \(\delta = 105^\circ\).

a) \(\beta = 100^\circ\) and \(\delta = 100^\circ\) b) \(\alpha = 75^\circ\), \(\beta = 85^\circ\), \(\gamma = 95^\circ\), and \(\delta = 105^\circ\)

- What is the sum of the interior angles of a triangle? - Express all three angles in terms of \(\alpha\). - How is a triangle with one \(90^\circ\) angle classified?

1. Use the triangle angle sum: \(\alpha + \beta + \gamma = 180^\circ\). 2. Substitute the given relationships: \(\alpha + 2\alpha + 3\alpha = 180^\circ\). 3. Combine like terms and solve: \(6\alpha = 180^\circ\), so \(\alpha = 30^\circ\). 4. Then \(\beta = 2 \cdot 30^\circ = 60^\circ\) and \(\gamma = 3 \cdot 30^\circ = 90^\circ\). 5. Since one angle measures \(90^\circ\), the triangle is a right triangle.

\(\alpha = 30^\circ\), \(\beta = 60^\circ\), and \(\gamma = 90^\circ\). The triangle is a right triangle.

- Express every angle in terms of \(\alpha\). - How many copies of \(\alpha\) make up the full \(180^\circ\) angle sum? - What conditions must three angle measures satisfy to form a triangle?

1. The relationships are \(\beta = 3\alpha\) and \(\gamma = 2\beta = 6\alpha\). 2. Use the triangle angle sum: \(\alpha + \beta + \gamma = 180^\circ\). 3. Substitute: \(\alpha + 3\alpha + 6\alpha = 180^\circ\), so \(10\alpha = 180^\circ\). 4. Therefore, \(\alpha = 18^\circ\), \(\beta = 3 \cdot 18^\circ = 54^\circ\), and \(\gamma = 2 \cdot 54^\circ = 108^\circ\). 5. The angles are all positive and total \(180^\circ\), so such a triangle does exist.

\(\alpha = 18^\circ\), \(\beta = 54^\circ\), and \(\gamma = 108^\circ\). Yes, the triangle exists.

- Check the relevant angle sum in each statement. - Recall the definitions of right and obtuse angles. - What is true about the base angles of an isosceles triangle? - How are an interior angle and its adjacent exterior angle related?

1. For part a, \(35^\circ + 45^\circ + 110^\circ = 190^\circ\). Since the total is not \(180^\circ\), the statement is false. 2. For part b, a right angle measures \(90^\circ\), while an obtuse angle is greater than \(90^\circ\). A triangle cannot contain both because their sum would already exceed \(180^\circ\). The statement is true. 3. For part c, the two base angles measure \(70^\circ\) each. The vertex angle is \(180^\circ - 70^\circ - 70^\circ = 40^\circ\). The statement is true. 4. For part d, each interior angle of an equilateral triangle is \(60^\circ\). Its adjacent exterior angle is \(180^\circ - 60^\circ = 120^\circ\), which is twice \(60^\circ\). The statement is true.

a) False; the angles total \(190^\circ\). b) True; a right angle and an obtuse angle cannot both fit in a triangle. c) True; the vertex angle is \(40^\circ\). d) True; the interior angle is \(60^\circ\) and the adjacent exterior angle is \(120^\circ\).

- What is true about the base angles of an isosceles triangle? - Express the vertex angle in terms of one base angle. - How many copies of the base angle make up the full \(180^\circ\) angle sum?

1. The base angles of an isosceles triangle are congruent, so \(\alpha = \beta\). 2. The vertex angle satisfies \(\gamma = 4\alpha\). 3. Use the triangle angle sum: \(\alpha + \beta + \gamma = 180^\circ\). 4. Substitute the relationships: \(\alpha + \alpha + 4\alpha = 180^\circ\), so \(6\alpha = 180^\circ\). 5. Thus \(\alpha = 30^\circ\), \(\beta = 30^\circ\), and \(\gamma = 4 \cdot 30^\circ = 120^\circ\).

\(\alpha = 30^\circ\), \(\beta = 30^\circ\), and \(\gamma = 120^\circ\).

- What is the interior angle sum of a quadrilateral? - Write \(\delta\) in terms of \(\gamma\). - How much of the \(360^\circ\) total remains after subtracting the known angles?

1. The interior angles of a quadrilateral total \(360^\circ\). 2. Substitute the known angles into \(\alpha + \beta + \gamma + \delta = 360^\circ\): \(85^\circ + 110^\circ + \gamma + \delta = 360^\circ\). 3. Since \(\delta = \gamma - 15^\circ\), write \(195^\circ + \gamma + (\gamma - 15^\circ) = 360^\circ\). 4. Simplify: \(180^\circ + 2\gamma = 360^\circ\), so \(2\gamma = 180^\circ\) and \(\gamma = 90^\circ\). 5. Then \(\delta = 90^\circ - 15^\circ = 75^\circ\).

\(\gamma = 90^\circ\) and \(\delta = 75^\circ\).

- How is the interior angle sum related to the number of triangles formed from one vertex? - What does “regular” tell you about the angle measures? - Write an equation with the number of sides as the unknown.

1. Use the interior angle sum formula \(S = (n - 2) \cdot 180^\circ\): \(1440^\circ = (n - 2) \cdot 180^\circ\). 2. Divide by \(180^\circ\): \(8 = n - 2\), so \(n = 10\). The polygon is a decagon. 3. Since the polygon is regular, its interior angles are congruent. Each angle measures \(1440^\circ \div 10 = 144^\circ\).

a) The polygon has \(10\) sides. b) Each interior angle measures \(144^\circ\).

- How many degrees does the robot turn during one complete trip? - Divide the total turn by the turn at each vertex. - What kind of number must the number of sides be?

1. After one complete trip around the polygon, the robot has turned a total of \(360^\circ\). 2. For part a, the number of sides is \(360^\circ \div 45^\circ = 8\), so the robot traces a regular octagon. 3. For part b, \(360^\circ \div 50^\circ = 7.2\). 4. A polygon must have a whole-number count of sides, so a regular polygon with a \(50^\circ\) exterior angle is not possible.

a) A regular octagon b) No. The calculation gives \(7.2\) sides, which is not possible for a polygon.

- First find the interior angle sum of the second polygon. - Use the polygon interior angle sum formula. - How much does the angle sum increase when one side is added?

1. The second polygon has an interior angle sum of \(1260^\circ + 360^\circ = 1620^\circ\). 2. Use \((n - 2) \cdot 180^\circ = 1620^\circ\). 3. Divide by \(180^\circ\): \(n - 2 = 9\), so \(n = 11\). 4. Equivalently, increasing the angle sum by \(360^\circ\) adds \(360 \div 180 = 2\) sides.

The second polygon has \(11\) sides.

- What is the sum of \(\alpha\), \(\beta\), and \(\gamma\)? - Express all three angles in terms of \(\alpha\). - What equation represents the three angle measures? - Does one of the calculated angles have a special measure?

1. Use the triangle angle sum and the given relationships: \(\alpha + 3\alpha + 2\alpha = 180^\circ\). 2. Combine like terms: \(6\alpha = 180^\circ\). 3. Solve for \(\alpha\): \(\alpha = 180^\circ \div 6 = 30^\circ\). 4. Find the other angles: \(\beta = 3 \cdot 30^\circ = 90^\circ\) and \(\gamma = 2 \cdot 30^\circ = 60^\circ\). 5. Because the triangle has a \(90^\circ\) angle, it is a right triangle.

1. \(\alpha = 30^\circ\), \(\beta = 90^\circ\), \(\gamma = 60^\circ\) 2. The triangle is a right triangle.

- Recall the definition of an obtuse angle. - Use the triangle angle sum to find a missing angle. - What must be true about two angles of an isosceles triangle? - Check each result against the \(180^\circ\) angle sum.

1. For a), a right angle measures \(90^\circ\), and an obtuse angle measures more than \(90^\circ\). Their sum would be greater than \(180^\circ\), which is impossible in a triangle. 2. For b), the third angle measures \(180^\circ - 85^\circ = 95^\circ\). Because \(95^\circ > 90^\circ\), the triangle is obtuse. 3. For c), the third angle would measure \(180^\circ - (90^\circ + 60^\circ) = 30^\circ\). The three angles would be \(30^\circ\), \(60^\circ\), and \(90^\circ\), with no two congruent angles. Therefore, the triangle could not be isosceles.

a) A right angle and an obtuse angle would have a sum greater than \(180^\circ\). b) The triangle is obtuse because the third angle measures \(95^\circ\). c) No. The angles would be \(30^\circ\), \(60^\circ\), and \(90^\circ\), so no two angles would be congruent.

- What does “regular” tell you about the exterior angles? - Divide the full \(360^\circ\) turn into \(8\) equal parts. - How are adjacent interior and exterior angles related? - Multiply one interior angle by the number of sides.

1. A regular octagon has \(8\) congruent exterior angles. One exterior angle measures \(360^\circ \div 8 = 45^\circ\). 2. An interior angle and its adjacent exterior angle are supplementary, so one interior angle measures \(180^\circ - 45^\circ = 135^\circ\). 3. The interior angle sum is \(8 \cdot 135^\circ = 1080^\circ\).

a) \(45^\circ\) b) \(135^\circ\) c) \(1080^\circ\)

- Which angles are congruent in an isosceles triangle? - How does knowing one angle help you find the other two? - Use the \(180^\circ\) triangle angle sum.

1. For a), the base angles have a sum of \(180^\circ - 110^\circ = 70^\circ\). Since they are congruent, \(\alpha = \beta = 70^\circ \div 2 = 35^\circ\). 2. For b), the other base angle is also \(\beta = 35^\circ\). The vertex angle is \(\gamma = 180^\circ - (35^\circ + 35^\circ) = 110^\circ\). 3. For c), a right vertex angle is possible. The remaining \(90^\circ\) is divided equally between the base angles, so \(\alpha = \beta = 90^\circ \div 2 = 45^\circ\).

a) \(\alpha = \beta = 35^\circ\) b) \(\beta = 35^\circ\) and \(\gamma = 110^\circ\) c) Yes. The base angles are each \(45^\circ\).

- How are an interior angle and its adjacent exterior angle related? - What is the sum of the three interior angles of a triangle? - A quick sketch may help you place the exterior angles.

1. An interior angle and its adjacent exterior angle form a linear pair and total \(180^\circ\). 2. Therefore, \(\alpha = 180^\circ - 125^\circ = 55^\circ\). 3. Similarly, \(\beta = 180^\circ - 110^\circ = 70^\circ\). 4. Use the triangle angle sum: \(\gamma = 180^\circ - 55^\circ - 70^\circ = 55^\circ\).

\(\alpha = 55^\circ\), \(\beta = 70^\circ\), and \(\gamma = 55^\circ\).

- Work with the two triangles separately first. - How are the full angles at \(A\) and \(C\) built from the smaller angles? - Relate the angle sums of the two triangles to the quadrilateral’s angle sum.

1. In triangle \(ABC\), \(\angle ABC = 180^\circ - 35^\circ - 45^\circ = 100^\circ\). 2. In triangle \(ADC\), \(\angle ADC = 180^\circ - 40^\circ - 60^\circ = 80^\circ\). 3. The full angle at \(A\) is \(35^\circ + 40^\circ = 75^\circ\), and the full angle at \(C\) is \(45^\circ + 60^\circ = 105^\circ\). 4. The quadrilateral angle sum is \(75^\circ + 100^\circ + 105^\circ + 80^\circ = 360^\circ\). Equivalently, the two triangles contribute \(180^\circ + 180^\circ = 360^\circ\).

a) \(\angle ABC = 100^\circ\) and \(\angle ADC = 80^\circ\) b) \(360^\circ\)

- What is the sum of the interior angles of a triangle? - What type of triangle has three congruent angles? - What is true about the side lengths of an equilateral triangle?

1. Use the triangle angle sum: \(\gamma = 180^\circ - 60^\circ - 60^\circ = 60^\circ\). 2. All three interior angles measure \(60^\circ\), so the triangle is equiangular and therefore equilateral. 3. All sides of an equilateral triangle are congruent. Since \(c = 8\,\text{cm}\), it follows that \(a = 8\,\text{cm}\) and \(b = 8\,\text{cm}\).

a) \(\gamma = 60^\circ\) b) \(a = 8\,\text{cm}\) and \(b = 8\,\text{cm}\). The triangle is equilateral because an equiangular triangle has three congruent sides.

- What is the interior angle sum of a quadrilateral? - Express every angle in terms of \(\alpha\). - Write one equation using the full angle sum. - Check that your four results total \(360^\circ\).

1. The interior angles of a quadrilateral total \(360^\circ\). 2. Let \(\alpha = x\). Then \(\beta = x\), \(\gamma = x + 30^\circ\), and \(\delta = 2x + 10^\circ\). 3. Write the equation \(x + x + (x + 30^\circ) + (2x + 10^\circ) = 360^\circ\). 4. Simplify: \(5x + 40^\circ = 360^\circ\). 5. Solve: \(5x = 320^\circ\), so \(x = 64^\circ\). 6. Therefore, \(\alpha = 64^\circ\), \(\beta = 64^\circ\), \(\gamma = 94^\circ\), and \(\delta = 2 \cdot 64^\circ + 10^\circ = 138^\circ\).

\(\alpha = 64^\circ\), \(\beta = 64^\circ\), \(\gamma = 94^\circ\), and \(\delta = 138^\circ\).

- Find how much of a full \(360^\circ\) turn is left. - Find how much the angle exceeds \(180^\circ\).

1. The remaining angle in a full turn is \(360^\circ-215^\circ=145^\circ\). 2. The amount beyond a straight angle is \(215^\circ-180^\circ=35^\circ\).

The remainder of a full turn is \(145^\circ\), and the amount beyond a straight angle is \(35^\circ\).

- Choose three angles less than \(90^\circ\). - Subtract their sum from \(360^\circ\). - Check whether the fourth angle is a valid interior angle.

1. Choose three acute angles of \(80^\circ\) each. Their sum is \(3 \cdot 80^\circ=240^\circ\). 2. The fourth angle must be \(360^\circ-240^\circ=120^\circ\). 3. The measures \(80^\circ\), \(80^\circ\), \(80^\circ\), and \(120^\circ\) sum to \(360^\circ\), so such a quadrilateral is possible.

No. One example has interior angles \(80^\circ\), \(80^\circ\), \(80^\circ\), and \(120^\circ\).

- At \({:}30\), the hour hand is halfway between two hour marks. - Each hour mark is \(30^\circ\) apart. - Find each hand’s position from \(12\), then subtract.

1. At half past an hour, the minute hand is at \(180^\circ\) from \(12\), and the hour hand is halfway between two hour marks. 2. At \(4{:}30\), the hour hand is at \(4 \cdot 30^\circ+15^\circ=135^\circ\). The angle is \(180^\circ-135^\circ=45^\circ\). 3. At \(10{:}30\), the hour hand is at \(10 \cdot 30^\circ+15^\circ=315^\circ\). The smaller angle is \(315^\circ-180^\circ=135^\circ\).

a) \(45^\circ\) b) \(135^\circ\)

- Each hour section measures \(30^\circ\). - Follow the specified counterclockwise direction. - Compare each angle measure with \(90^\circ\), \(180^\circ\), and \(360^\circ\).

1. Each hour section measures \(360^\circ \div 12=30^\circ\). 2. At \(2{:}00\), the counterclockwise angle is \(2 \cdot 30^\circ=60^\circ\), which is acute. 3. At \(3{:}00\), the angle is \(3 \cdot 30^\circ=90^\circ\), which is right. 4. At \(5{:}00\), the angle is \(5 \cdot 30^\circ=150^\circ\), which is obtuse. 5. At \(6{:}00\), the angle is \(6 \cdot 30^\circ=180^\circ\), which is straight. 6. At \(8{:}00\), the counterclockwise angle is \(8 \cdot 30^\circ=240^\circ\), which is reflex.

a) Acute b) Right c) Obtuse d) Straight e) Reflex

- Substitute each number of sides into the given formula. - Solve the formula for \(n\) in part b. - What does “regular” tell you about how the total angle sum is divided?

1. For a hexagon, \(S = (6 - 2) \cdot 180^\circ = 720^\circ\). 2. For an octagon, \(S = (8 - 2) \cdot 180^\circ = 1080^\circ\). 3. For part b, solve \(1260^\circ = (n - 2) \cdot 180^\circ\). Dividing by \(180^\circ\) gives \(7 = n - 2\), so \(n = 9\). 4. Since the polygon is regular, one interior angle measures \(1260^\circ \div 9 = 140^\circ\).

a) Hexagon: \(720^\circ\); octagon: \(1080^\circ\) b) \(n = 9\) c) \(140^\circ\)

- How are an interior angle and its adjacent exterior angle related? - Use the \(360^\circ\) sum of the exterior angles. - The number of sides must be a whole number.

1. For part a, \(\alpha = 4\beta\) and \(\alpha + \beta = 180^\circ\). 2. Substitute: \(4\beta + \beta = 180^\circ\), so \(5\beta = 180^\circ\) and \(\beta = 36^\circ\). 3. The number of sides is \(n = 360^\circ \div 36^\circ = 10\). 4. For part b, an interior angle of \(175^\circ\) has an exterior angle of \(180^\circ - 175^\circ = 5^\circ\). 5. Then \(n = 360^\circ \div 5^\circ = 72\). Since \(72\) is a whole number, the polygon exists.

a) \(n = 10\) b) Yes. The polygon has \(72\) sides.

- Divide the full interior angle sum by the number of congruent angles. - Rearrange the equation in part b so that the terms containing \(n\) are on one side. - Can \(\frac{n - 2}{n}\) ever equal or exceed \(1\)?

1. For a regular hexagon, \(S = (6 - 2) \cdot 180^\circ = 720^\circ\), so one angle is \(720^\circ \div 6 = 120^\circ\). 2. For a regular dodecagon, \(S = (12 - 2) \cdot 180^\circ = 1800^\circ\), so one angle is \(1800^\circ \div 12 = 150^\circ\). 3. For part b, solve \(\frac{(n - 2) \cdot 180^\circ}{n} = 144^\circ\). This gives \(180n - 360 = 144n\), so \(36n = 360\) and \(n = 10\). 4. No. In \(\alpha = \frac{n - 2}{n} \cdot 180^\circ\), the factor \(\frac{n - 2}{n}\) is always less than \(1\), so \(\alpha < 180^\circ\). Geometrically, an angle of \(180^\circ\) would not form a vertex.

a) Hexagon: \(120^\circ\); dodecagon: \(150^\circ\) b) \(n = 10\) c) No. A regular polygon’s interior angle is always less than \(180^\circ\).

- What is true about the base angles of an isosceles triangle? - What is the sum of the interior angles of a triangle? - Express the vertex angle in terms of one base angle.

1. Let each base angle be \(\alpha\), and let the vertex angle be \(\gamma\). 2. The relationship is \(\gamma = 0.5\alpha\). 3. Use the triangle angle sum: \(\alpha + \alpha + \gamma = 180^\circ\). 4. Substitute: \(2\alpha + 0.5\alpha = 180^\circ\), so \(2.5\alpha = 180^\circ\). 5. Therefore, \(\alpha = 72^\circ\), and \(\gamma = 0.5 \cdot 72^\circ = 36^\circ\).

The base angles each measure \(72^\circ\), and the vertex angle measures \(36^\circ\).

- First find the interior angle at \(B\). How is it related to the exterior angle? - What is the sum of the interior angles of a triangle? - Once two interior angles are known, subtract them from \(180^\circ\).

1. The interior angle at \(B\) and the \(120^\circ\) exterior angle form a linear pair, so the interior angle at \(B\) is \(180^\circ - 120^\circ = 60^\circ\). 2. The interior angles of a triangle total \(180^\circ\). 3. Therefore, \(\gamma = 180^\circ - 50^\circ - 60^\circ = 70^\circ\).

\(\gamma = 70^\circ\)

- What is the measure of a straight angle? - How does an angle bisector divide \(\alpha\)? - What sum do angles in a linear pair have?

1. A straight angle measures \(180^\circ\). 2. Since \(w_\alpha\) bisects \(\alpha\), the angle between the rightward ray of \(g\) and \(w_\alpha\) is \(\frac{\alpha}{2}\). 3. That angle and \(\delta\) form a linear pair, so \(180^\circ-\frac{\alpha}{2}=145^\circ\). 4. Thus, \(\frac{\alpha}{2}=35^\circ\), so \(\alpha=70^\circ\). 5. Since \(\alpha\) and \(\beta\) form a linear pair, \(\beta=180^\circ-70^\circ=110^\circ\).

\(\alpha=70^\circ\) and \(\beta=110^\circ\).

- Represent \(\beta\) as two equal copies of \(\alpha\). - Write an equation for the total angle measure.

1. The angle measures satisfy \(\alpha+\beta=135^\circ\). 2. Because \(\beta\) is twice \(\alpha\), \(\beta=2\alpha\). 3. Substitute to get \(\alpha+2\alpha=135^\circ\), so \(3\alpha=135^\circ\). 4. Therefore, \(\alpha=45^\circ\), and \(\beta=2(45^\circ)=90^\circ\).

\(\alpha=45^\circ\) and \(\beta=90^\circ\)

- What does an angle bisector do? - First find the angle between \(a\) and \(w_1\). - Which angle does \(w_2\) bisect? - Add the two adjacent parts that form \(\beta\).

1. Since \(w_1\) bisects the \(74^{\circ}\) angle, \(\angle(a,w_1) = \angle(w_1,b) = 37^{\circ}\). 2. Line \(w_2\) bisects the \(37^{\circ}\) angle between \(w_1\) and \(b\), so \(\angle(w_1,w_2) = 18.5^{\circ}\). 3. Therefore, \(\beta = 37^{\circ} + 18.5^{\circ} = 55.5^{\circ}\).

\(\beta = 55.5^{\circ}\)

- How are an interior angle and its adjacent exterior angle related? - What is the interior angle sum of a quadrilateral? - Find the angles connected to the given exterior angles before using the quadrilateral angle sum.

1. The \(70^\circ\) exterior angle and interior angle \(\gamma\) form a linear pair, so \(\gamma = 180^\circ - 70^\circ = 110^\circ\). 2. The \(75^\circ\) interior angle and exterior angle \(\beta\) form a linear pair, so \(\beta = 180^\circ - 75^\circ = 105^\circ\). 3. The interior angles of a quadrilateral total \(360^\circ\). Therefore, \(\alpha + 110^\circ + 75^\circ + 90^\circ = 360^\circ\). 4. Thus \(\alpha = 360^\circ - 275^\circ = 85^\circ\).

\(\alpha = 85^\circ\), \(\beta = 105^\circ\), and \(\gamma = 110^\circ\).

- What relationship do \(\angle ADB\) and \(\angle ADC\) have? - Use the angle sum in triangle \(ABD\). - How does an angle bisector relate \(\angle DAB\) and \(\angle CAD\)?

1. Angles \(\angle ADB\) and \(\angle ADC\) form a linear pair, so \(\angle ADB=180^\circ-75^\circ=105^\circ\). 2. In triangle \(ABD\), \(\angle DAB=180^\circ-105^\circ-40^\circ=35^\circ\). 3. Since \(\overline{AD}\) bisects \(\angle BAC\), \(\alpha=\angle CAD=\angle DAB=35^\circ\). 4. Therefore, \(\angle BAC=2\cdot35^\circ=70^\circ\). 5. In triangle \(ABC\), \(\gamma=180^\circ-70^\circ-40^\circ=70^\circ\).

\(\alpha=35^\circ\) and \(\gamma=70^\circ\).

- What is the measure of a straight angle? - How are an interior angle and its adjacent exterior angle related? - What is the sum of the interior angles of a triangle?

1. The \(125^\circ\) exterior angle and interior angle \(\alpha\) form a linear pair, so \(\alpha = 180^\circ - 125^\circ = 55^\circ\). 2. The interior angles of a triangle total \(180^\circ\). 3. Therefore, \(\gamma = 180^\circ - 55^\circ - 65^\circ = 60^\circ\).

\(\alpha = 55^\circ\) and \(\gamma = 60^\circ\).

- How are an interior angle and its adjacent exterior angle related? - What is the interior angle sum of a pentagon? - Subtract all known interior angles from the pentagon’s total.

1. At \(A\), \(\alpha\) and the \(75^\circ\) exterior angle form a linear pair, so \(\alpha = 180^\circ - 75^\circ = 105^\circ\). 2. At \(C\), \(\gamma\) and the \(85^\circ\) exterior angle form a linear pair, so \(\gamma = 180^\circ - 85^\circ = 95^\circ\). 3. The interior angle sum of a pentagon is \((5 - 2) \cdot 180^\circ = 540^\circ\). 4. Therefore, \(\epsilon = 540^\circ - 105^\circ - 115^\circ - 95^\circ - 130^\circ = 95^\circ\).

\(\alpha = 105^\circ\), \(\gamma = 95^\circ\), and \(\epsilon = 95^\circ\).

- Which angles are congruent because of the kite’s symmetry? - What is the interior angle sum of a quadrilateral? - Write one equation for all four angles.

1. Since \(AC\) is the symmetry axis, the angles at \(B\) and \(D\) are congruent, so \(\beta = \delta\). 2. The interior angles of a quadrilateral total \(360^\circ\). 3. Substitute the known angles: \(38^\circ + \beta + 82^\circ + \beta = 360^\circ\). 4. Simplify: \(120^\circ + 2\beta = 360^\circ\), so \(2\beta = 240^\circ\) and \(\beta = 120^\circ\). 5. Therefore, \(\delta = 120^\circ\).

\(\beta = 120^\circ\) and \(\delta = 120^\circ\).

- What is true about the base angles of an isosceles triangle? - What is the sum of the interior angles of a triangle? - Write an equation using the two congruent base angles.

1. Since \(AC = BC\), the base angles at \(A\) and \(B\) are congruent, so \(\alpha = \beta\). 2. Use the triangle angle sum: \(\alpha + \beta + 54^\circ = 180^\circ\). 3. Substitute \(\beta = \alpha\): \(2\alpha + 54^\circ = 180^\circ\). 4. Then \(2\alpha = 126^\circ\), so \(\alpha = 63^\circ\).

\(\alpha = 63^\circ\)

- How are an interior angle and its adjacent exterior angle related? - What is the sum of the interior angles of a triangle? - Find \(\gamma\) before finding \(\beta\).

1. The exterior angle at \(C\) and interior angle \(\gamma\) form a linear pair. 2. Therefore, \(\gamma = 180^\circ - 115^\circ = 65^\circ\). 3. Use the triangle angle sum: \(45^\circ + \beta + 65^\circ = 180^\circ\). 4. Thus \(\beta = 180^\circ - 45^\circ - 65^\circ = 70^\circ\).

\(\beta = 70^\circ\) and \(\gamma = 65^\circ\).

- What does it mean that \(\overline{AD}\) is an angle bisector? - How are \(\angle ADB\) and \(\angle ADC\) related? - First use triangle \(ABD\), then use the angle sum in triangle \(ABC\).

1. Angles \(\angle ADB\) and \(\angle ADC\) form a linear pair, so \(\angle ADB=180^\circ-95^\circ=85^\circ\). 2. In triangle \(ABD\), one half of \(\alpha\) is \(\frac{\alpha}{2}=180^\circ-70^\circ-85^\circ=25^\circ\). 3. Since \(\overline{AD}\) bisects \(\alpha\), the full angle is \(\alpha=2\cdot25^\circ=50^\circ\). 4. In triangle \(ABC\), \(\gamma=180^\circ-50^\circ-70^\circ=60^\circ\).

\(\alpha=50^\circ\) and \(\gamma=60^\circ\).

- How are an interior angle and its adjacent exterior angle related? - What is the interior angle sum of a quadrilateral? - Find \(\beta\) first, then use the quadrilateral angle sum.

1. The \(115^\circ\) exterior angle and interior angle \(\beta\) form a linear pair, so \(\beta = 180^\circ - 115^\circ = 65^\circ\). 2. The interior angles of a quadrilateral total \(360^\circ\). 3. Substitute the known values: \(75^\circ + 65^\circ + \gamma + 110^\circ = 360^\circ\). 4. Therefore, \(\gamma = 360^\circ - 250^\circ = 110^\circ\).

\(\beta = 65^\circ\) and \(\gamma = 110^\circ\).

- How is each exterior angle related to its adjacent interior angle? - Find the two unmarked interior angles first. - What is the sum of the interior angles of a triangle?

1. At \(A\), the \(130^\circ\) exterior angle and interior angle \(\alpha\) form a linear pair, so \(\alpha = 180^\circ - 130^\circ = 50^\circ\). 2. At \(C\), the \(120^\circ\) exterior angle and interior angle \(\gamma\) form a linear pair, so \(\gamma = 180^\circ - 120^\circ = 60^\circ\). 3. Use the triangle angle sum: \(50^\circ + \beta + 60^\circ = 180^\circ\). 4. Therefore, \(\beta = 70^\circ\).

\(\beta = 70^\circ\)

- How many sides does the polygon have? - What formula gives the sum of the interior angles of an \(n\)-gon? - After finding the total, how can you use the four given angles to find the missing angle?

1. The sum of the interior angles of an \(n\)-gon is \((n - 2) \cdot 180^\circ\). 2. A pentagon has \(5\) sides, so its interior angles total \((5 - 2) \cdot 180^\circ = 540^\circ\). 3. The four known angles total \(100^\circ + 110^\circ + 120^\circ + 95^\circ = 425^\circ\). 4. Therefore, \(\varphi = 540^\circ - 425^\circ = 115^\circ\).

\(\varphi = 115^\circ\)

- Use the triangle angle-sum theorem. - Use the fact that \(AD\) bisects angle \(A\). - Work with the smaller triangles. - Recall the relationship between a linear pair of angles.

1. The angles of triangle \(ABC\) sum to \(180^{\circ}\), so angle \(A\) is \(180^{\circ} - 70^{\circ} - 60^{\circ} = 50^{\circ}\). 2. Since \(AD\) bisects angle \(A\), \(\alpha = 50^{\circ} \div 2 = 25^{\circ}\). 3. In triangle \(ABD\), \(\beta = 180^{\circ} - 70^{\circ} - 25^{\circ} = 85^{\circ}\). 4. Angles \(\beta\) and \(\gamma\) form a linear pair, so \(\gamma = 180^{\circ} - 85^{\circ} = 95^{\circ}\).

\(\alpha = 25^{\circ}\), \(\beta = 85^{\circ}\), and \(\gamma = 95^{\circ}\)

- What special angle relationship follows from the kite's line of symmetry? - Which diagonal is the line of symmetry? - What does the symmetry tell you about the angles at \(B\) and \(D\)? - What is the sum of the interior angles of a quadrilateral? - How can you use that sum to find the final unknown angle?

1. Because \(AC\) is the line of symmetry, vertices \(B\) and \(D\) correspond. Their angle measures are equal, so \(\delta = 120^\circ\). 2. The interior angles of a quadrilateral total \(360^\circ\). 3. Substitute the known measures: \(50^\circ + 120^\circ + \gamma + 120^\circ = 360^\circ\). 4. Therefore, \(\gamma = 360^\circ - 290^\circ = 70^\circ\).

\(\gamma = 70^\circ\) and \(\delta = 120^\circ\)

- What sum do the three adjacent angles have? - Build the angle between the bisectors from half of \(\alpha\), all of \(\beta\), and half of \(\gamma\). - For part b, use \(\alpha+\gamma=120^\circ\) with \(\gamma=2\alpha\).

1. Since \(\alpha+\beta+\gamma=180^\circ\) and \(\beta=60^\circ\), \(\alpha+\gamma=120^\circ\). 2. The angle between the two bisectors is \(\frac{\alpha}{2}+\beta+\frac{\gamma}{2}\). 3. Therefore, it measures \(\frac{\alpha+\gamma}{2}+60^\circ=\frac{120^\circ}{2}+60^\circ=120^\circ\). 4. For part b, substitute \(\gamma=2\alpha\) into \(\alpha+\gamma=120^\circ\): \(\alpha+2\alpha=120^\circ\). 5. Thus, \(3\alpha=120^\circ\), so \(\alpha=40^\circ\) and \(\gamma=80^\circ\).

a) \(120^\circ\). b) \(\alpha=40^\circ\) and \(\gamma=80^\circ\).

- How does an angle bisector divide the \(50^\circ\) angle? - Which smaller triangle contains \(\delta\)? - Use the triangle angle sum in \(ADC\).

1. Since \(\overline{AD}\) bisects \(\alpha=50^\circ\), \(\angle CAD=50^\circ\div2=25^\circ\). 2. In triangle \(ADC\), the angles sum to \(180^\circ\). 3. Therefore, \(25^\circ+70^\circ+\delta=180^\circ\). 4. Thus, \(\delta=180^\circ-95^\circ=85^\circ\).

\(\delta=85^\circ\).

- How are an interior angle and its adjacent exterior angle related? - What is the sum of the interior angles of a quadrilateral? - Can you first find the interior angle at \(C\)? - Once three interior angles are known, how can you find the fourth?

1. First find the interior angle at \(C\). An interior angle and its adjacent exterior angle form a linear pair, so the interior angle is \(180^\circ - 105^\circ = 75^\circ\). 2. The interior angles of a quadrilateral total \(360^\circ\). 3. Therefore, \(\beta = 360^\circ - (85^\circ + 75^\circ + 90^\circ) = 110^\circ\).

\(\beta = 110^\circ\)

- Can you work with the two smaller triangles separately? - Which angles form a linear pair at point \(D\)? - What is the sum of the interior angles of any triangle? - Try finding a missing angle in triangle \(ADC\) or triangle \(BCD\) first.

1. Because \(A\), \(D\), and \(B\) are collinear, \(\angle ADC\) and \(\angle BDC\) form a linear pair. Thus, \(\angle ADC = 180^\circ - 80^\circ = 100^\circ\). 2. In triangle \(ADC\), \(\delta_1 = 180^\circ - 55^\circ - 100^\circ = 25^\circ\). 3. In triangle \(BCD\), \(\delta_2 = 180^\circ - 80^\circ - 40^\circ = 60^\circ\).

\(\delta_1 = 25^\circ\) and \(\delta_2 = 60^\circ\)

- How are \(\beta\) and the \(120^\circ\) angle related? - What is the sum of angles that form a straight line? - What figure is formed by the two slanted lines and line \(h\)? - How can the triangle angle sum determine the angle at the top?

1. Angle \(\beta\) and the given \(120^\circ\) angle form a linear pair. Therefore, \(\beta = 180^\circ - 120^\circ = 60^\circ\). 2. The two transversals and line \(h\) form a triangle whose base angles are \(65^\circ\) and \(60^\circ\). 3. Using the triangle angle sum, \(\gamma = 180^\circ - 65^\circ - 60^\circ = 55^\circ\).

\(\beta = 60^\circ\) and \(\gamma = 55^\circ\)

- Find the interior angles at \(A\), \(B\), and \(C\) first. - What is true about vertical angles? Use this at \(A\). - How can the exterior angle at \(C\) determine the interior angle there? - Use the interior angle sum of a quadrilateral to find \(\delta\).

1. The interior angle at \(A\) is vertical to the given \(120^\circ\) angle, so it measures \(120^\circ\). 2. The interior angle at \(B\) is \(70^\circ\). 3. The interior angle at \(C\) and the given \(75^\circ\) exterior angle form a linear pair, so the interior angle at \(C\) is \(180^\circ - 75^\circ = 105^\circ\). 4. The interior angles of a quadrilateral total \(360^\circ\). Therefore, \(\delta = 360^\circ - (120^\circ + 70^\circ + 105^\circ) = 65^\circ\).

\(\delta = 65^\circ\)

- What is the sum of the interior angles of a quadrilateral? - After subtracting the two known angles, what sum remains for \(\gamma\) and \(\delta\)? - If one angle is twice the other, into how many equal parts can you divide their combined measure?

1. The interior angles of a quadrilateral total \(360^\circ\), so \(75^\circ + 105^\circ + \gamma + \delta = 360^\circ\). 2. Therefore, \(\gamma + \delta = 180^\circ\). 3. Because \(\delta = 2\gamma\), substitute to get \(\gamma + 2\gamma = 180^\circ\). 4. Thus, \(3\gamma = 180^\circ\), so \(\gamma = 60^\circ\). Then \(\delta = 2 \cdot 60^\circ = 120^\circ\).

\(\gamma = 60^\circ\) and \(\delta = 120^\circ\)

- What is the sum of the interior angles of a triangle? - How are adjacent angles on a straight line related? - Start with the triangle on the left to find the angle at \(D\). - Use its supplement in the triangle on the right.

1. In triangle \(ADC\), \(40^\circ + 30^\circ + \delta = 180^\circ\). Therefore, \(\delta = 110^\circ\). 2. Because \(A\), \(D\), and \(B\) are collinear, \(\delta\) and \(\angle BDC\) form a linear pair. Thus, \(\angle BDC = 180^\circ - 110^\circ = 70^\circ\). 3. In triangle \(BDC\), \(70^\circ + 55^\circ + \gamma_2 = 180^\circ\). Therefore, \(\gamma_2 = 55^\circ\).

\(\delta = 110^\circ\) and \(\gamma_2 = 55^\circ\)

- First find \(\beta\). - Divide each angle by \(2\). - Which two half-angles combine to form \(\delta\)?

1. Since \(\alpha\) and \(\beta\) form a linear pair, \(\beta=180^\circ-74^\circ=106^\circ\). 2. The bisectors create angles of \(74^\circ\div2=37^\circ\) and \(106^\circ\div2=53^\circ\). 3. The angle between the bisectors is \(\delta=37^\circ+53^\circ=90^\circ\).

\(\delta=90^\circ\).

- Can you first find the interior angle of the triangle at \(B\)? - What is the sum of the interior angles of a triangle? - Which angle relationships follow from the parallel lines?

1. The \(130^\circ\) angle at \(B\) and the interior angle of triangle \(ABC\) at \(B\) form a linear pair. Therefore, the interior angle at \(B\) is \(180^\circ - 130^\circ = 50^\circ\). 2. The interior angles of triangle \(ABC\) total \(180^\circ\). Thus, \(\gamma = 180^\circ - 40^\circ - 50^\circ = 90^\circ\).

\(\gamma = 90^\circ\)

- First find the interior angle adjacent to the \(135^\circ\) exterior angle. - What is the sum of the interior angles of a triangle?

1. The right interior angle and the given \(135^\circ\) exterior angle form a linear pair. Therefore, the right interior angle is \(180^\circ - 135^\circ = 45^\circ\). 2. Using the triangle angle sum, \(\alpha = 180^\circ - 50^\circ - 45^\circ = 85^\circ\).

\(\alpha = 85^\circ\)

- What is the sum of the interior angles of a triangle? - How are angles in a linear pair related? - What is true about vertical angles?

1. The interior angle adjacent to the \(115^\circ\) exterior angle is \(180^\circ - 115^\circ = 65^\circ\). 2. The interior angle at the top vertex is vertical to the marked \(60^\circ\) angle, so it also measures \(60^\circ\). 3. Using the triangle angle sum, \(\alpha = 180^\circ - 65^\circ - 60^\circ = 55^\circ\).

\(\alpha = 55^\circ\)

- What is the sum of the interior angles of a quadrilateral? - Use linear-pair and vertical-angle relationships to find the interior angles at \(C\) and \(D\). - Subtract the three known interior angles from \(360^\circ\).

1. The interior angle at \(D\) and the marked \(110^\circ\) exterior angle form a linear pair, so the interior angle at \(D\) is \(70^\circ\). 2. The interior angle at \(C\) is vertical to the marked \(110^\circ\) angle, so it measures \(110^\circ\). 3. The interior angles of a quadrilateral total \(360^\circ\). Therefore, \(\alpha = 360^\circ - (80^\circ + 110^\circ + 70^\circ) = 100^\circ\).

\(\alpha = 100^\circ\)

- Does the \(360^\circ\) interior-angle sum still apply to a concave quadrilateral? - The missing interior angle is a reflex angle, so it is greater than \(180^\circ\).

1. The interior angles of any quadrilateral, including a concave quadrilateral, total \(360^\circ\). 2. The three given angles total \(30^\circ + 40^\circ + 30^\circ = 100^\circ\). 3. Therefore, \(\delta = 360^\circ - 100^\circ = 260^\circ\).

\(\delta = 260^\circ\)

- Use one variable for the two congruent unknown angles. - How much of the \(360^\circ\) total remains after subtracting the known angles?

1. The interior angles of a quadrilateral total \(360^\circ\). 2. Since \(\alpha = \gamma\), write \(2\alpha + 30^\circ + 280^\circ = 360^\circ\). 3. Then \(2\alpha = 50^\circ\), so \(\alpha = 25^\circ\).

\(\alpha = 25^\circ\)

- What does the angle bisector tell you about the two angles at \(A\)? - Use the angle sum in triangle \(ABC\). - Then use the angle sum in triangle \(ADC\).

1. Since \(\overline{AD}\) bisects \(\angle BAC\), the full angle at \(A\) is \(2\cdot32^\circ=64^\circ\). 2. In triangle \(ABC\), \(\gamma=180^\circ-64^\circ-44^\circ=72^\circ\). 3. In triangle \(ADC\), \(\delta=180^\circ-32^\circ-72^\circ=76^\circ\).

\(\gamma=72^\circ\) and \(\delta=76^\circ\).

- First find each interior angle that forms a linear pair with a given exterior angle. - What does a \(90^\circ\) exterior angle imply about its adjacent interior angle? - Then use the interior-angle sum of a quadrilateral.

1. At \(A\), \(\alpha = 180^\circ - 110^\circ = 70^\circ\). 2. The interior angle at \(B\) is \(180^\circ - 85^\circ = 95^\circ\). 3. At \(C\), \(\gamma = 180^\circ - 90^\circ = 90^\circ\). 4. The interior angles of a quadrilateral total \(360^\circ\). Therefore, \(\delta = 360^\circ - (70^\circ + 95^\circ + 90^\circ) = 105^\circ\).

\(\alpha = 70^\circ\), \(\gamma = 90^\circ\), and \(\delta = 105^\circ\)

- Which angles are congruent in each isosceles triangle? - Use the triangle angle sum. - How are adjacent angles on line \(AEB\) related? - Work from the left triangle to the right triangle.

1. In isosceles triangle \(ADE\), the base angles at \(A\) and \(D\) are congruent. Therefore, \(\angle ADE = 35^\circ\). 2. Then \(\angle AED = 180^\circ - 35^\circ - 35^\circ = 110^\circ\). 3. Because \(A\), \(E\), and \(B\) are collinear, \(\angle DEB = 180^\circ - 110^\circ = 70^\circ\). 4. In isosceles triangle \(EBD\), the base angles at \(D\) and \(B\) are congruent. Thus, \(70^\circ + 2\varepsilon = 180^\circ\), so \(\varepsilon = 55^\circ\).

\(\varepsilon = 55^\circ\)

- The two opposite rays of a line form a \(180^\circ\) angle. - Express the direction of both rays from the same side of the line. - Then find the difference between those direction angles.

1. Ray \(u\) forms an angle of \(180^\circ - 115^\circ = 65^\circ\) with the right-pointing part of line \(g\). 2. Ray \(v\) forms a \(110^\circ\) angle with that same part of the line. 3. Subtract the direction angles: \(\alpha = 110^\circ - 65^\circ = 45^\circ\).

\(\alpha = 45^\circ\)

- An angle bisector creates two congruent angles. - First halve \(120^\circ\), then halve one of the resulting angles. - Combine the two smaller angles between \(c\) and \(e\).

1. Since \(d\) bisects the \(120^\circ\) angle, the angles between \(a\) and \(d\), and between \(d\) and \(c\), each measure \(60^\circ\). 2. Since \(e\) bisects the \(60^\circ\) angle between \(a\) and \(d\), the angle between \(d\) and \(e\) measures \(30^\circ\). 3. The angle between \(c\) and \(e\) is \(60^\circ+30^\circ=90^\circ\).

The angle between rays \(c\) and \(e\) is \(90^\circ\).

- First find the angle supplementary to \(126^\circ\). - Bisect that angle. - Then use a linear pair to find the requested angle.

1. The angle between \(m\) and the rightward ray of \(g\) is supplementary to \(126^\circ\), so it measures \(180^\circ-126^\circ=54^\circ\). 2. Since \(k\) bisects this angle, each half measures \(54^\circ\div2=27^\circ\). 3. The angle between \(k\) and the leftward ray of \(g\) measures \(180^\circ-27^\circ=153^\circ\).

The angle measures \(153^\circ\).

- What is the sum of the interior angles of a pentagon? - How does symmetry relate the angles at \(C\) and \(E\)?

1. The sum of the interior angles of a pentagon is \((5 - 2) \cdot 180^\circ = 540^\circ\). 2. The three known angles total \(90^\circ + 90^\circ + 110^\circ = 290^\circ\). 3. The two remaining angles total \(540^\circ - 290^\circ = 250^\circ\). 4. By symmetry, the two remaining angles are congruent, so \(\gamma = 250^\circ \div 2 = 125^\circ\).

Each angle \(\gamma\) measures \(125^\circ\).

- How do you find the sum of the interior angles of a hexagon? - After subtracting the known angles, divide the remaining total equally among the other four angles.

1. The sum of the interior angles of a hexagon is \((6 - 2) \cdot 180^\circ = 720^\circ\). 2. Subtract the two right angles: \(720^\circ - 2 \cdot 90^\circ = 540^\circ\). 3. Divide the remaining total among the four congruent angles: \(\gamma = 540^\circ \div 4 = 135^\circ\).

\(\gamma = 135^\circ\)

- Identify the upper and lower triangles. - What is the relationship between the opposite angles at the central intersection? - Use the triangle angle sum in each triangle. - Lines \(g\) and \(h\) do not need to be parallel.

1. In the lower triangle, the angle at the intersection of \(s\) and \(t\) is \(180^\circ - 45^\circ - 45^\circ = 90^\circ\). 2. The angle at the same intersection in the upper triangle is vertical to that angle, so it also measures \(90^\circ\). 3. Use the angle sum of the upper triangle: \(\delta = 180^\circ - 90^\circ - 35^\circ = 55^\circ\).

\(\delta = 55^\circ\)

- Use the angle sum of the large triangle to find its missing acute angle. - Then use the angle sum in the smaller triangle containing the \(100^\circ\) angle. - The two marked angles together form the right angle at \(C\).

1. In triangle \(ABC\), \(\angle B = 180^\circ - 90^\circ - 25^\circ = 65^\circ\). 2. In triangle \(BDC\), \(\alpha = 180^\circ - 65^\circ - 100^\circ = 15^\circ\). 3. The right angle at \(C\) is divided into \(\alpha\) and \(\beta\), so \(\beta = 90^\circ - 15^\circ = 75^\circ\).

\(\alpha = 15^\circ\) and \(\beta = 75^\circ\)

- Start with the smaller triangle containing two given angles. - How are the two angles at point \(D\) related? - Then use the angle sum in the other smaller triangle.

1. In triangle \(BDC\), \(\beta = 180^\circ - 50^\circ - 35^\circ = 95^\circ\). 2. The two angles at \(D\) form a linear pair, so \(\angle ADC = 180^\circ - 95^\circ = 85^\circ\). 3. In triangle \(ADC\), \(\alpha = 180^\circ - 85^\circ - 25^\circ = 70^\circ\).

\(\alpha = 70^\circ\) and \(\beta = 95^\circ\)

- Find a smaller triangle in which two angle measures are known. - Use the linear pair at \(D\) to move from one smaller triangle to the other. - Apply the triangle angle sum to each smaller triangle.

1. In triangle \(ADC\), \(\alpha = 180^\circ - 45^\circ - 85^\circ = 50^\circ\). 2. The two angles at \(D\) form a linear pair, so \(\angle BDC = 180^\circ - 85^\circ = 95^\circ\). 3. In triangle \(BDC\), \(\beta = 180^\circ - 95^\circ - 55^\circ = 30^\circ\).

\(\alpha = 50^\circ\) and \(\beta = 30^\circ\)

- Use linear pairs to find the two base angles. - Then use the triangle angle sum. - Classify the triangle using its equal angles and its right angle.

1. Each base interior angle forms a linear pair with a \(135^\circ\) exterior angle, so \(\angle A = \angle B = 180^\circ - 135^\circ = 45^\circ\). 2. The third interior angle is \(\angle C = 180^\circ - 45^\circ - 45^\circ = 90^\circ\). 3. Because two angles are congruent, the triangle is isosceles. Because one angle is a right angle, it is also a right triangle.

The interior angles are \(45^\circ\), \(45^\circ\), and \(90^\circ\). The triangle is an isosceles right triangle.

- Divide the figure into two isosceles triangles. - Which angles in each smaller triangle are congruent? - The full angle \(\beta\) is the sum of two angles at \(B\).

1. Since \(AD = BD\), triangle \(ABD\) is isosceles, so \(\angle ABD = \alpha = 25^\circ\). 2. The exterior angle \(\angle BDC\) of triangle \(ABD\) is \(25^\circ + 25^\circ = 50^\circ\). 3. Since \(BD = CD\), triangle \(BCD\) is isosceles. Its two base angles each measure \((180^\circ - 50^\circ) \div 2 = 65^\circ\), so \(\gamma = 65^\circ\). 4. The full angle at \(B\) is \(\beta = 25^\circ + 65^\circ = 90^\circ\).

\(\beta = 90^\circ\) and \(\gamma = 65^\circ\)

- Identify the two isosceles triangles in the figure. - Use the exterior angle at \(M\) to connect their angle measures. - Then use the angle sum in triangle \(BCM\).

1. Since \(AM = MB\), triangle \(ABM\) is isosceles, so \(\angle ABM = \angle BAM = 38^\circ\). 2. The exterior angle \(\angle BMC\) of triangle \(ABM\) is \(38^\circ + 38^\circ = 76^\circ\). 3. Since \(MB = MC\), triangle \(BCM\) is isosceles. Its base angles are congruent, so \(\gamma = (180^\circ - 76^\circ) \div 2 = 52^\circ\).

\(\gamma = 52^\circ\)

- What angles of triangle \(ABI\) are created by the two angle bisectors? - Use the angle sum in triangle \(ABI\). - Then use the angle sum in triangle \(ABC\).

1. In triangle \(ABI\), the angles at \(A\) and \(B\) are half of the corresponding angles of triangle \(ABC\). Therefore, \(\frac{\angle A}{2}+\frac{\angle B}{2}+130^\circ=180^\circ\). 2. Thus, \(\frac{\angle A}{2}+\frac{\angle B}{2}=50^\circ\), so \(\angle A+\angle B=100^\circ\). 3. Using the angle sum of triangle \(ABC\), \(\angle C=180^\circ-100^\circ=80^\circ\).

\(\angle C=80^\circ\).

- First find the interior angle adjacent to the \(108^\circ\) exterior angle. - Then use the triangle angle sum and compare the three interior angles.

1. The interior angle at \(A\) is \(180^\circ - 108^\circ = 72^\circ\). 2. Use the triangle angle sum: \(\angle C = 180^\circ - 72^\circ - 36^\circ = 72^\circ\). 3. Since \(\angle A = \angle C = 72^\circ\), the triangle is isosceles with base \(AC\).

\(\angle C = 72^\circ\). The triangle is isosceles because \(\angle A = \angle C\).

- Convert each exterior angle to its adjacent interior angle. - What does equality of two interior angles tell you about the triangle?

1. Each corresponding interior angle forms a linear pair with a \(125^\circ\) exterior angle, so \(\angle A = \angle C = 180^\circ - 125^\circ = 55^\circ\). 2. The remaining angle is \(\angle B = 180^\circ - 55^\circ - 55^\circ = 70^\circ\). 3. Since two interior angles are congruent, the triangle is isosceles.

\(\angle A = 55^\circ\), \(\angle B = 70^\circ\), and \(\angle C = 55^\circ\). The triangle is isosceles.

- First find the interior angle adjacent to the exterior angle at \(B\).

1. The interior angle at \(B\) forms a linear pair with the exterior angle, so \(\angle B = 180^\circ - 100^\circ = 80^\circ\). 2. Use the triangle angle sum: \(\alpha = 180^\circ - 80^\circ - 38^\circ = 62^\circ\).

\(\alpha = 62^\circ\)

- First find the interior angles at \(A\) and \(B\).

1. Convert the exterior angles to interior angles: \(\angle A = 180^\circ - 130^\circ = 50^\circ\) and \(\angle B = 180^\circ - 110^\circ = 70^\circ\). 2. Use the triangle angle sum: \(\gamma = 180^\circ - 50^\circ - 70^\circ = 60^\circ\).

\(\gamma = 60^\circ\)

- First find the interior angle at \(B\).

1. The interior angle at \(B\) is \(180^\circ - 140^\circ = 40^\circ\). 2. The two acute angles of a right triangle sum to \(90^\circ\), so \(\alpha = 90^\circ - 40^\circ = 50^\circ\).

\(\alpha = 50^\circ\)

- What is the relationship between vertical angles?

1. Vertical angles are congruent, so \(\alpha = 42^\circ\). 2. Use the triangle angle sum: \(\beta = 180^\circ - 42^\circ - 68^\circ = 70^\circ\).

\(\beta = 70^\circ\)

- Use the exterior angle to find one base angle, then use the isosceles triangle relationship.

1. The interior angle at \(A\) is \(180^\circ - 135^\circ = 45^\circ\). 2. The base angles are congruent, so \(\angle B = 45^\circ\). 3. The vertex angle is \(\angle C = 180^\circ - 45^\circ - 45^\circ = 90^\circ\).

\(\angle A = 45^\circ\), \(\angle B = 45^\circ\), and \(\angle C = 90^\circ\)

- Use the marked \(65^\circ\) angle and first find the interior angle adjacent to \(110^\circ\). - Then apply the triangle angle sum.

1. The left interior angle of the triangle above the horizontal line measures \(65^\circ\). 2. The right interior angle forms a linear pair with the \(110^\circ\) angle, so it measures \(180^\circ - 110^\circ = 70^\circ\). 3. Use the triangle angle sum: \(x = 180^\circ - 65^\circ - 70^\circ = 45^\circ\).

\(x = 45^\circ\)

- Recall the sum of two consecutive angles in a parallelogram. - Represent the smaller angle with one variable. - Write an equation using the four-to-one relationship. - Recall the relationship between opposite angles in a parallelogram.

1. Consecutive angles in a parallelogram are supplementary, so \(\angle K + \angle L = 180^\circ\). 2. Let \(\angle L = x\). Then \(\angle K = 4x\), so \(x + 4x = 180^\circ\). 3. Solve: \(5x = 180^\circ\), so \(x = 36^\circ\). Therefore, \(\angle L = 36^\circ\) and \(\angle K = 4 \cdot 36^\circ = 144^\circ\). 4. Opposite angles in a parallelogram are congruent. Thus, \(\angle M = 144^\circ\) and \(\angle N = 36^\circ\).

\(\angle K = 144^\circ\), \(\angle L = 36^\circ\), \(\angle M = 144^\circ\), and \(\angle N = 36^\circ\)

- What is the sum of adjacent angles in a parallelogram? - What does an angle bisector do? - Use the angle sum of triangle \(ABP\).

1. Adjacent angles in a parallelogram are supplementary, so \(\alpha + \beta = 180^{\circ}\). 2. The angle bisectors create angles of \(\frac{\alpha}{2}\) and \(\frac{\beta}{2}\) in triangle \(ABP\). 3. Thus \(\frac{\alpha}{2} + \frac{\beta}{2} = \frac{180^{\circ}}{2} = 90^{\circ}\). 4. By the triangle angle-sum theorem, \(\angle APB = 180^{\circ} - 90^{\circ} = 90^{\circ}\).

\(\angle APB = 90^{\circ}\)

- Use the isosceles triangle formed by \(A\), \(B\), and \(M\). - Then apply the angle relationships in a parallelogram.

1. Since \(AB = AM\), triangle \(ABM\) is isosceles, so \(\angle ABM = \angle BMA = 70^\circ\). 2. The angle at \(A\) is \(180^\circ - 70^\circ - 70^\circ = 40^\circ\). 3. Opposite angles of a parallelogram are congruent, so \(\angle C = 40^\circ\). 4. Consecutive angles are supplementary, so \(\angle B = \angle D = 180^\circ - 40^\circ = 140^\circ\).

The interior angles are \(40^\circ\), \(140^\circ\), \(40^\circ\), and \(140^\circ\).

- Use the isosceles triangle formed by \(A\), \(D\), and \(C\). - Then use the angle relationship created by diagonal \(AC\) crossing the parallel bases.

1. Since \(AD = DC\), triangle \(ADC\) is isosceles. 2. Its base angles are congruent, so \(\angle DAC = \angle DCA = (180^\circ - 110^\circ) \div 2 = 35^\circ\). 3. Since \(AB \parallel CD\), \(\angle BAC\) and \(\angle DCA\) are alternate interior angles. Therefore, \(\angle BAC = 35^\circ\).

\(\angle BAC = 35^\circ\)

- First find the exterior angle corresponding to the \(140^\circ\) interior angle. - The exterior angles of any polygon total \(360^\circ\).

1. Each exterior angle is supplementary to its interior angle, so it measures \(180^\circ - 140^\circ = 40^\circ\). 2. The exterior angles of a polygon total \(360^\circ\), so the number of vertices is \(n = \frac{360^\circ}{40^\circ} = 9\).

The polygon has \(9\) vertices.

- The minute hand moves \(6^\circ\) per minute. - The hour hand moves \(30^\circ\) per hour, or \(0.5^\circ\) per minute. - Find both positions from \(12\), then take their difference.

1. At \(15\) minutes, the minute hand is \(15 \cdot 6^\circ = 90^\circ\) clockwise from \(12\). 2. At \(2{:}15\), the hour hand is \(2 \cdot 30^\circ + 15 \cdot 0.5^\circ = 67.5^\circ\) clockwise from \(12\). 3. The smaller angle between the hands is \(|90^\circ - 67.5^\circ| = 22.5^\circ\).

\(22.5^\circ\)

- Each hour mark represents \(30^\circ\). - Find each hand’s position relative to \(12\). - Remember that the hour hand moves between hour marks as the minutes pass.

1. At \(20\) minutes, the minute hand is \(20 \cdot 6^\circ = 120^\circ\) clockwise from \(12\). 2. At \(8{:}20\), the hour hand is \(8 \cdot 30^\circ + 20 \cdot 0.5^\circ = 250^\circ\) clockwise from \(12\). 3. The difference is \(250^\circ - 120^\circ = 130^\circ\), which is the smaller angle.

\(130^\circ\)

- First find the measure of the supplementary angle. - Bisect each angle. - Add the two half-angles that form the requested angle.

1. Since \(\alpha\) and \(\beta\) are supplementary, \(\beta = 180^{\circ} - 124^{\circ} = 56^{\circ}\). 2. Their half-angles are \(124^{\circ} \div 2 = 62^{\circ}\) and \(56^{\circ} \div 2 = 28^{\circ}\). 3. The angle between the bisectors is \(\delta = 62^{\circ} + 28^{\circ} = 90^{\circ}\).

The angle between the two angle bisectors is \(90^{\circ}\).

- If you know half of \(\alpha\), how can you find the full angle? - What is the sum of a linear pair? - How does \(w_\beta\) divide \(\beta\)?

1. Since \(w_\alpha\) bisects \(\alpha\), \(\alpha=2\cdot37^\circ=74^\circ\). 2. The angles form a linear pair, so \(\beta=180^\circ-74^\circ=106^\circ\). 3. Since \(w_\beta\) bisects \(\beta\), the angle between \(s\) and \(w_\beta\) is \(106^\circ\div2=53^\circ\).

\(\alpha=74^\circ\), \(\beta=106^\circ\), and the angle between \(s\) and \(w_\beta\) is \(53^\circ\).

- What is true about the two angles not on the line of symmetry? - What is the sum of the interior angles of a quadrilateral? - Write an equation with the unknown angle.

1. In a kite with symmetry line \(AC\), the opposite angles at \(B\) and \(D\) are congruent, so \(\beta = \delta\). 2. The interior angles of a quadrilateral sum to \(360^{\circ}\), so \(80^{\circ} + \beta + 40^{\circ} + \beta = 360^{\circ}\). 3. Combine terms: \(120^{\circ} + 2\beta = 360^{\circ}\). 4. Solve: \(2\beta = 240^{\circ}\), so \(\beta = 120^{\circ}\).

The measure of \(\beta\) is \(120^{\circ}\).

- Use the angle sum of a triangle to find \(\angle DAB\). - Recall the relationship between opposite angles of a parallelogram.

1. The angle sum of triangle \(ABD\) is \(180^\circ\), so \(\angle DAB = 180^\circ - 42^\circ - 38^\circ = 100^\circ\). 2. Opposite angles of a parallelogram are congruent, so \(\angle BCD = \angle DAB = 100^\circ\).

\(\angle BCD = 100^\circ\)

- Use the equal side lengths in the square and the equilateral triangle. - Identify the isosceles triangles formed by point \(M\). - Write \(\angle AMC\) as a sum of two angles at \(M\).

1. Since \(AD = AM\), triangle \(ADM\) is isosceles. Also, \(\angle DAM = 90^\circ - 60^\circ = 30^\circ\), so \(\angle AMD = 75^\circ\). 2. Similarly, \(BC = BM\) and \(\angle CBM = 30^\circ\), so \(\angle BMC = 75^\circ\). 3. Therefore, \(\angle AMC = \angle AMB + \angle BMC = 60^\circ + 75^\circ = 135^\circ\).

\(\angle AMC = 135^\circ\)

- Sketch the square and attach an equilateral triangle along one side. - What are the angle measures in a square and an equilateral triangle? - Which sides must be congruent because of the definitions of a square and an equilateral triangle? - What is true about the base angles of an isosceles triangle?

1. At \(C\), the square contributes a \(90^\circ\) angle and the equilateral triangle contributes a \(60^\circ\) angle. Therefore, \(m\angle DCE = 90^\circ + 60^\circ = 150^\circ\). 2. In the square, \(DC = BC\). In the equilateral triangle, \(BC = CE\). Therefore, \(DC = CE\), so \(\triangle DCE\) is isosceles. 3. The two base angles of \(\triangle DCE\) are congruent. Their sum is \(180^\circ - 150^\circ = 30^\circ\), so \(m\angle EDC = 30^\circ \div 2 = 15^\circ\).

a) \(m\angle DCE = 150^\circ\) b) Since \(DC = BC\) and \(BC = CE\), it follows that \(DC = CE\). Thus, \(\triangle DCE\) is isosceles. c) \(m\angle EDC = 15^\circ\)

- Substitute the given angle into the formula for one interior angle of a regular polygon. - Double the number of sides found in part a. - Calculate the new angle and compare it with twice \(150^\circ\).

1. For a regular polygon, \(\alpha = \frac{(n - 2) \cdot 180^\circ}{n}\). 2. Substitute \(\alpha = 150^\circ\): \(150n = 180(n - 2)\). 3. Solve: \(150n = 180n - 360\), so \(30n = 360\) and \(n = 12\). 4. Doubling the number of sides gives \(24\) sides. The new interior angle is \(\frac{(24 - 2) \cdot 180^\circ}{24} = 165^\circ\). 5. Since \(165^\circ \ne 2 \cdot 150^\circ\), the claim is false.

a) The polygon has \(12\) sides. b) The claim is false. A regular polygon with \(24\) sides has interior angles of \(165^\circ\), not \(300^\circ\).

- How are an interior angle and its adjacent exterior angle related? - Express the vertex angle in terms of a base angle. - Use one variable for both congruent base angles.

1. Let each base angle be \(\alpha\), and let the vertex angle be \(\gamma\). 2. The exterior angle at a base vertex is \(180^\circ - \alpha\). 3. The given relationship is \(180^\circ - \alpha = \gamma + 45^\circ\). 4. Since the triangle is isosceles, \(\gamma = 180^\circ - 2\alpha\). 5. Substitute: \(180^\circ - \alpha = 180^\circ - 2\alpha + 45^\circ\). 6. Solving gives \(\alpha = 45^\circ\). 7. Then \(\gamma = 180^\circ - 2 \cdot 45^\circ = 90^\circ\).

The interior angles are \(45^\circ\), \(45^\circ\), and \(90^\circ\).

- The hour hand does not remain fixed between whole hours. - Find how far the hour hand moves in \(15\) minutes. - Subtract that movement from \(90^\circ\).

1. During the first \(15\) minutes after \(12{:}00\), the hour hand moves partway from \(12\) toward \(1\), so it is no longer at \(12\). 2. The minute hand is at \(90^\circ\) from \(12\). 3. The hour hand moves \(30^\circ\) in \(60\) minutes, so in \(15\) minutes it moves \(30^\circ \cdot \frac{15}{60}=7.5^\circ\). 4. The angle between the hands is \(90^\circ-7.5^\circ=82.5^\circ\).

a) The hour hand moves continuously and has already moved toward \(1\). b) \(82.5^\circ\)

- What does the converse of the isosceles triangle theorem say about two congruent angles? - Use the triangle angle sum to find the missing angle in \(\triangle ABD\). - For part b), first find \(m\angle C\) in \(\triangle ABC\). - Compare the angles in \(\triangle ADC\).

1. In \(\triangle ABD\), \(m\angle BAD = 36^\circ\) and \(m\angle ABD = 72^\circ\). Therefore, \(m\angle ADB = 180^\circ - 36^\circ - 72^\circ = 72^\circ\). Since \(\angle ABD \cong \angle ADB\), the opposite sides are congruent: \(AD = AB\). 2. In \(\triangle ABC\), \(m\angle C = 180^\circ - 72^\circ - 72^\circ = 36^\circ\). In \(\triangle ADC\), \(m\angle DAC = 36^\circ\) and \(m\angle ACD = 36^\circ\). Since these angles are congruent, their opposite sides are congruent: \(AD = CD\).

a) \(\triangle ABD\) is isosceles because \(\angle ABD \cong \angle ADB\). Its congruent sides are \(AB\) and \(AD\). b) \(\triangle ADC\) is isosceles because \(\angle DAC \cong \angle ACD\). Its congruent sides are \(AD\) and \(CD\).

- What must \(\beta\) be so that \(\alpha + \beta = 120^\circ\)? - Find half of each smaller angle. - Add the two parts between the bisectors. - Compare the result with the original \(120^\circ\) angle.

1. The second angle is \(\beta = 120^\circ - 40^\circ = 80^\circ\). 2. The bisectors create angles of \(\frac{40^\circ}{2} = 20^\circ\) and \(\frac{80^\circ}{2} = 40^\circ\) next to the dividing ray. 3. Therefore, \(\gamma = 20^\circ + 40^\circ = 60^\circ\). 4. In general, the angle between the internal bisectors of two adjacent angles is half the sum of those angles. Thus, it is half the original angle.

\(\gamma = 60^\circ\). In general, \(\gamma\) is one-half of the original angle.

- Mark the congruent sides and identify the base angles in each isosceles triangle. - Use the linear-pair relationship at \(D\) to move from one triangle to the other. - Use the \(180^\circ\) triangle angle sum. - Add the two angle parts at \(B\) to classify the large triangle.

1. In isosceles triangle \(BDC\), the base angles at \(B\) and \(C\) are congruent. Therefore, \(\gamma = \frac{180^\circ - 116^\circ}{2} = 32^\circ\). 2. Because \(D\) lies on \(AC\), \(\angle ADB\) and \(\angle BDC\) form a linear pair. Thus, \(\angle ADB = 180^\circ - 116^\circ = 64^\circ\). 3. In isosceles triangle \(ABD\), the base angles at \(A\) and \(B\) are congruent. Therefore, \(\alpha = \frac{180^\circ - 64^\circ}{2} = 58^\circ\). 4. The full angle at \(B\) is \(\angle ABC = 58^\circ + 32^\circ = 90^\circ\). 5. Therefore, triangle \(ABC\) is a right triangle.

a) \(\alpha = 58^\circ\) and \(\gamma = 32^\circ\) b) \(\angle ABC = 90^\circ\), so \(ABC\) is a right triangle.

- Start with the linear pair containing \(\beta\). - Use the angle sum of the left triangle. - Then use a linear pair and vertical angles to transfer the information to the right triangle. - Finish with the angle sum of the right triangle.

1. The interior angle adjacent to \(\beta\) in the left triangle is \(180^\circ - 115^\circ = 65^\circ\). 2. The third angle of the left triangle is \(180^\circ - 65^\circ - 45^\circ = 70^\circ\). 3. The adjacent angle in the right triangle forms a linear pair with that \(70^\circ\) angle, so it measures \(180^\circ - 70^\circ = 110^\circ\). 4. The top angle of the right triangle is vertical to \(\gamma\), so it measures \(40^\circ\). 5. Use the triangle angle sum: \(x = 180^\circ - 110^\circ - 40^\circ = 30^\circ\).

\(x = 30^\circ\)

- Use \(AD = DC\) to relate angles in triangle \(ADC\). - Use alternate interior angles along the parallel bases. - Use the right triangle formed by \(A\), \(B\), and \(C\).

1. Let \(\angle BAC = \alpha\). Since \(AB \parallel CD\), alternate interior angles give \(\angle DCA = \alpha\). 2. Since \(AD = DC\), triangle \(ADC\) is isosceles, so \(\angle DAC = \angle DCA = \alpha\). Therefore, \(\angle A = 2\alpha\). 3. An isosceles trapezoid has congruent base angles, so \(\angle B = 2\alpha\). 4. Triangle \(ABC\) is right at \(C\), so \(\alpha + 2\alpha = 90^\circ\). Thus, \(\alpha = 30^\circ\). 5. Therefore, \(\angle A = \angle B = 60^\circ\), and the supplementary upper angles satisfy \(\angle C = \angle D = 120^\circ\).

\(\angle A = 60^\circ\), \(\angle B = 60^\circ\), \(\angle C = 120^\circ\), and \(\angle D = 120^\circ\)

- Use the side lengths of the square and the equilateral triangle to identify isosceles triangles. - Find the angles at \(P\) in triangles \(ABP\) and \(DCP\). - The angles around point \(P\) total \(360^\circ\).

1. Because \(ABCD\) is a square and triangle \(BCP\) is equilateral, \(AB = BP\) and \(CD = CP\). 2. At \(B\), \(\angle ABP = 90^\circ - 60^\circ = 30^\circ\). 3. Triangle \(ABP\) is isosceles, so \(\angle BPA = \frac{180^\circ - 30^\circ}{2} = 75^\circ\). 4. In the same way, \(\angle CPD = 75^\circ\). 5. The angles around \(P\) total \(360^\circ\), so \(\angle APD = 360^\circ - 75^\circ - 60^\circ - 75^\circ = 150^\circ\).

\(\angle APD = 150^\circ\)

- Compare the side lengths of the square and the equilateral triangle. - Find the vertex angle of isosceles triangle \(ABP\). - Use the symmetry of the figure and the \(60^\circ\) angle at \(P\).

1. Since \(AB = BC = BP\), triangle \(ABP\) is isosceles. 2. Its vertex angle is \(\angle ABP = 90^\circ + 60^\circ = 150^\circ\), so \(\angle BPA = \frac{180^\circ - 150^\circ}{2} = 15^\circ\). 3. Similarly, triangle \(DCP\) is isosceles and \(\angle DPC = 15^\circ\). 4. Since \(\angle BPC = 60^\circ\), \(\angle APD = 60^\circ - 15^\circ - 15^\circ = 30^\circ\).

\(\angle APD = 30^\circ\)

- A rhombus has four congruent sides. - What is true about the diagonals of a rhombus? - How do the diagonals divide the vertex angles? - Use the triangle angle-sum theorem.

1. In a rhombus, all four sides are congruent. In \(\triangle PQS\), \(PQ = PS\), and the included angle at \(P\) is \(60^{\circ}\). The other two angles are each \((180^{\circ} - 60^{\circ}) \div 2 = 60^{\circ}\), so \(\triangle PQS\) is equilateral and acute. 2. In \(\triangle PQR\), \(PQ = QR\), so it is isosceles. The angle at \(Q\) is supplementary to the \(60^{\circ}\) angle at \(P\), so it is \(120^{\circ}\). Thus \(\triangle PQR\) is obtuse and isosceles. 3. The diagonals of a rhombus are perpendicular and bisect the vertex angles. In \(\triangle PQZ\), the angles are \(30^{\circ}\) at \(P\), \(60^{\circ}\) at \(Q\), and \(90^{\circ}\) at \(Z\). It is a scalene right triangle.

a) \(\triangle PQS\) is equilateral and acute. b) \(\triangle PQR\) is isosceles and obtuse. c) The angles are \(30^{\circ}\), \(60^{\circ}\), and \(90^{\circ}\); the triangle is scalene and right.