Free Math Worksheets

- What is true about the base angles of an isosceles triangle? - Use the triangle angle sum to find them. - Match each side with its opposite angle in the Law of Sines.

1. The base angles are equal: \(\alpha=\beta=\frac{180^\circ-36^\circ}{2}=72^\circ\). 2. Apply the Law of Sines: \(\frac{a}{\sin(72^\circ)}=\frac{8.4}{\sin(36^\circ)}\). 3. Solve: \(a=\frac{8.4\sin(72^\circ)}{\sin(36^\circ)}\approx 13.59\,\text{cm}\). 4. Since the triangle is isosceles, \(b=a\approx 13.59\,\text{cm}\).

The congruent sides are \(a\approx 13.59\,\text{cm}\) and \(b\approx 13.59\,\text{cm}\).

- Use the triangle angle sum first. - Which law relates each side to the sine of its opposite angle? - Match each side with the correct opposite angle.

1. Use the triangle angle sum: \(\gamma=180^\circ-42^\circ-75^\circ=63^\circ\). 2. By the Law of Sines, \(b=\frac{8.5\sin(75^\circ)}{\sin(42^\circ)}\approx 12.27\,\text{cm}\). 3. Similarly, \(c=\frac{8.5\sin(63^\circ)}{\sin(42^\circ)}\approx 11.32\,\text{cm}\).

\(\gamma=63^\circ\), \(b\approx 12.27\,\text{cm}\), and \(c\approx 11.32\,\text{cm}\).

- Find the third angle using the triangle angle sum. - Use the known side-angle pair in the Law of Sines. - Keep each side matched with its opposite angle.

1. Find the third angle: \(\gamma=180^\circ-40^\circ-60^\circ=80^\circ\). 2. By the Law of Sines, \(a=\frac{9\sin(40^\circ)}{\sin(80^\circ)}\approx 5.9\,\text{cm}\). 3. Similarly, \(b=\frac{9\sin(60^\circ)}{\sin(80^\circ)}\approx 7.9\,\text{cm}\).

\(\gamma=80^\circ\), \(a\approx 5.9\,\text{cm}\), and \(b\approx 7.9\,\text{cm}\).

- SSA is not automatically a congruence criterion. - Use the Law of Sines to find the angle opposite the \(4\,\text{cm}\) side. - Test both the inverse-sine angle and its supplement against the triangle angle sum.

1. For \(\triangle ABC\), apply the Law of Sines: \(\frac{\sin \beta}{4} = \frac{\sin 100^\circ}{7}\). Thus, \(\sin \beta = \frac{4\sin 100^\circ}{7} \approx 0.5627\). 2. One possible value is \(\beta \approx 34.2^\circ\). The supplementary value, approximately \(145.8^\circ\), is impossible because \(100^\circ + 145.8^\circ > 180^\circ\). Therefore, the data determine a unique triangle. 3. The same calculation applies to \(\triangle DEF\), so \(\epsilon \approx 34.2^\circ\), and its supplementary value is also impossible. 4. Both sets determine the same side lengths and angle measures, so the triangles are congruent.

Yes. Each SSA data set produces one unique triangle, so \(\triangle ABC\) and \(\triangle DEF\) are congruent. The second sine solution is impossible because the given \(100^\circ\) angle is already obtuse.

- Identify the angle between sides \(a\) and \(b\). - Use the sum and difference forms of the triangle inequality. - For SSA, solve for \(\sin\beta\) and check an angle and its supplement.

1. The sides \(a\) and \(b\) meet at vertex \(C\), so their included angle is \(\gamma\). Knowing \(\gamma\) would give SAS. 2. The triangle inequality gives \(|6 - 4| < c < 6 + 4\), so \(2\,\text{cm} < c < 10\,\text{cm}\). 3. For c), the Law of Sines gives \(\sin\beta = \frac{6\sin 30^\circ}{4} = 0.75\). 4. Thus, \(\beta \approx 48.6^\circ\) or \(\beta \approx 131.4^\circ\). Both angles can be combined with \(\alpha = 30^\circ\), so two noncongruent triangles are possible.

a) \(\gamma\) b) \(2\,\text{cm} < c < 10\,\text{cm}\) c) The SSA data produce two possible values of \(\beta\), so the triangle is not uniquely determined.

- For Lucas, check whether the given angle is included between the two sides. - For Mia, use the Law of Sines to find a possible value of \(\beta\). - Test whether the supplement of that angle can also fit in the triangle.

1. Lucas has sides \(a\) and \(c\) with their included angle \(\beta\). His triangle is uniquely determined by SAS. 2. Mia has SSA data. By the Law of Sines, \(\sin\beta = \frac{4.5\sin 55^\circ}{7} \approx 0.5266\). 3. One possible angle is \(\beta \approx 31.8^\circ\). Its supplement, \(148.2^\circ\), cannot be used because \(55^\circ + 148.2^\circ > 180^\circ\). 4. Therefore, Mia's measurements also determine exactly one triangle, but this conclusion comes from checking the SSA ambiguous case rather than applying a standard congruence criterion.

Lucas: One unique triangle by SAS. Mia: One unique triangle. The Law of Sines gives only one valid SSA solution.

- Find the third angle first. - Use the Law of Sines to determine \(c\). - Then use the area formula with two sides and their included angle.

1. Find the third angle: \(\beta=180^\circ-45^\circ-60^\circ=75^\circ\). 2. By the Law of Sines, \(\frac{c}{\sin(60^\circ)}=\frac{8}{\sin(75^\circ)}\), so \(c=\frac{8\sin(60^\circ)}{\sin(75^\circ)}\approx 7.17\,\text{cm}\). 3. Use the included angle \(\alpha\) between sides \(b\) and \(c\): \(A=\frac{1}{2}bc\sin(\alpha)\). 4. Therefore, \(A=\frac{1}{2}\cdot 8\cdot 7.1726\cdot \sin(45^\circ)\approx 20.29\,\text{cm}^2\).

a) \(c\approx 7.17\,\text{cm}\) b) The area is approximately \(20.29\,\text{cm}^2\).

- Find the third angle first. - Use the Law of Sines with the known baseline and its opposite angle. - Match the requested distance with its opposite angle.

1. Find the third angle: \(\gamma=180^\circ-42^\circ-105^\circ=33^\circ\). 2. The distance \(BP\) is opposite \(\alpha\). By the Law of Sines, \(\frac{BP}{\sin(42^\circ)}=\frac{80}{\sin(33^\circ)}\). 3. Therefore, \(BP=80\frac{\sin(42^\circ)}{\sin(33^\circ)}\approx 98.29\,\text{m}\).

The shelter is approximately \(98.29\,\text{m}\) from point \(B\).

- Find the two congruent base angles first. - Pair each side with its opposite angle in the Law of Sines. - Solve the proportion for the base.

1. The two base angles are congruent, so each is \(\frac{180^{\circ}-40^{\circ}}{2}=70^{\circ}\). 2. By the Law of Sines, \(\frac{c}{\sin(40^{\circ})}=\frac{10}{\sin(70^{\circ})}\). 3. Therefore, \(c=\frac{10\sin(40^{\circ})}{\sin(70^{\circ})}\approx 6.84\,\text{cm}\).

\(c \approx 6.84\,\text{cm}\)

- Use the triangle angle sum to find the third angle. - Match side \(a\) with angle \(\alpha\) and side \(c\) with angle \(\gamma\) in the Law of Sines.

1. The third angle is \(\gamma=180^{\circ}-45^{\circ}-105^{\circ}=30^{\circ}\). 2. By the Law of Sines, \(\frac{x}{\sin(45^{\circ})}=\frac{2}{\sin(30^{\circ})}\). 3. Thus, \(x=\frac{2\sin(45^{\circ})}{\sin(30^{\circ})}=2\sqrt{2}\approx 2.83\).

\(x=2\sqrt{2}\approx 2.83\)

- Identify which given side is opposite the known angle and which given side is not. - Apply the Law of Sines to find a possible value of \(\alpha\). - Check the supplementary angle as a second possibility. - Complete each possible triangle separately.

1. This is the ambiguous SSA case. Since \(a\sin(\gamma)<c<a\), two triangles are possible. 2. By the Law of Sines, \(\sin(\alpha)=\frac{a\sin(\gamma)}{c}=\frac{6.4\sin(35^\circ)}{4.5}\approx 0.81575\). 3. The first possible angle is \(\alpha_1=\sin^{-1}(0.81575)\approx 54.66^\circ\). The second is \(\alpha_2=180^\circ-54.66^\circ\approx 125.34^\circ\). 4. For Triangle 1, \(\beta_1=180^\circ-35^\circ-54.66^\circ\approx 90.34^\circ\), and \(b_1=\frac{4.5\sin(90.34^\circ)}{\sin(35^\circ)}\approx 7.85\,\text{cm}\). 5. For Triangle 2, \(\beta_2=180^\circ-35^\circ-125.34^\circ\approx 19.66^\circ\), and \(b_2=\frac{4.5\sin(19.66^\circ)}{\sin(35^\circ)}\approx 2.64\,\text{cm}\).

Two triangles are possible because the given information is the ambiguous SSA case and \(a\sin(\gamma)<c<a\). Triangle 1: \(\alpha_1\approx 54.66^\circ\), \(\beta_1\approx 90.34^\circ\), and \(b_1\approx 7.85\,\text{cm}\). Triangle 2: \(\alpha_2\approx 125.34^\circ\), \(\beta_2\approx 19.66^\circ\), and \(b_2\approx 2.64\,\text{cm}\).

- These are SSA cases, so check for the ambiguous case. - Use the Law of Sines to find the sine of the unknown angle. - A value greater than \(1\) is impossible, and a supplementary angle must still fit the triangle angle sum.

1. Case 1 gives \(b = 8\,\text{cm}\), \(a = 6\,\text{cm}\), and \(\beta = 50^\circ\). By the Law of Sines, \(\sin \alpha = \frac{6\sin 50^\circ}{8} \approx 0.5745\). 2. This gives \(\alpha \approx 35.1^\circ\). The supplementary possibility, \(144.9^\circ\), is impossible because \(144.9^\circ + 50^\circ > 180^\circ\). Therefore, Case 1 produces one triangle. 3. Case 2 gives \(a = 6\,\text{cm}\), \(b = 8\,\text{cm}\), and \(\alpha = 50^\circ\). The Law of Sines would require \(\sin \beta = \frac{8\sin 50^\circ}{6} \approx 1.0214\). 4. Since a sine value cannot exceed \(1\), Case 2 produces no triangle.

Case 1: One triangle Case 2: No triangle

- Recognize that the information forms an SSA case. - Use the Law of Sines to find the angle opposite the \(5\,\text{cm}\) side. - Test both inverse-sine possibilities with the \(100^\circ\) angle.

1. The data are \(b = 9\,\text{cm}\), \(c = 5\,\text{cm}\), and \(\beta = 100^\circ\), with the given angle opposite the longer side. 2. Use the Law of Sines: \(\sin \gamma = \frac{5\sin 100^\circ}{9} \approx 0.5471\). 3. One possible angle is \(\gamma \approx 33.2^\circ\). Its supplement, approximately \(146.8^\circ\), is impossible because \(100^\circ + 146.8^\circ > 180^\circ\). 4. Therefore, exactly one triangle satisfies the data. Lara’s conclusion is correct. Tom is incorrect because an obtuse angle does not have to be the included angle for the SSA data to determine a unique triangle.

Lara is correct: the data determine exactly one triangle. Tom’s claim is false; the Law of Sines shows that the only valid additional angle is approximately \(33.2^\circ\).

- Treat \(\triangle ABD\) as an SSA ambiguous case. - Use the Law of Sines and test both inverse-sine angles. - Once \(D\) is fixed, how many sides of \(BD\) can contain the SSS triangle \(BCD\)?

1. In \(\triangle ABD\), \(BD = 5\,\text{cm}\) is opposite \(\angle A = 30^\circ\), and \(AB = 8\,\text{cm}\). 2. By the Law of Sines, \(\sin \angle D = \frac{8\sin 30^\circ}{5} = 0.8\). Therefore, \(\angle D\) can be approximately \(53.1^\circ\) or \(126.9^\circ\). Both fit with \(\angle A = 30^\circ\), so there are two possible triangles \(ABD\) and two possible locations for \(D\). 3. For either location of \(D\), triangle \(BCD\) has side lengths \(4\,\text{cm}\), \(5\,\text{cm}\), and \(5\,\text{cm}\), so it is determined by SSS. 4. Triangle \(BCD\) can be attached on either side of \(\overline{BD}\). Thus, each of the two choices for \(D\) gives two choices for \(C\), for a total of \(2 \cdot 2 = 4\) quadrilaterals.

1. There are two possible locations for \(D\) because the SSA data for \(\triangle ABD\) produce two valid triangles. 2. There are \(4\) quadrilaterals: two choices for \(D\), and for each one, two choices for the side of \(BD\) containing \(C\).

- Write a Law of Sines equation for \(\sin\beta\). - Remember that an acute angle and its supplement have the same sine. - Check whether each possible angle leaves a positive measure for the third angle.

1. Case 1: By the Law of Sines, \(\sin \beta = \frac{b\sin\alpha}{a} = \frac{5\sin 30^\circ}{8} = 0.3125\). 2. One possible angle is \(\beta \approx 18.2^\circ\). Its supplement, \(161.8^\circ\), cannot be used because \(30^\circ + 161.8^\circ > 180^\circ\). Therefore, exactly one triangle is possible. 3. Case 2: \(\sin \beta = \frac{8\sin 30^\circ}{5} = 0.8\). 4. The two possible angles are \(\beta \approx 53.1^\circ\) and \(\beta \approx 126.9^\circ\). Each can be combined with \(\alpha = 30^\circ\), so two noncongruent triangles are possible.

Case 1: Exactly one triangle is determined. Case 2: Two noncongruent triangles are possible, so the triangle is not uniquely determined.

- What is \(\sin(90^\circ)\)? - Rearrange the proportion to isolate \(\sin(\alpha)\). - Equal angles have equal sine values.

1. Since \(\gamma=90^\circ\), \(\sin(\gamma)=1\). Therefore, \(\frac{a}{\sin(\alpha)}=\frac{c}{1}=c\), which rearranges to \(\sin(\alpha)=\frac{a}{c}\). 2. The Law of Sines is a general relationship. In a right triangle, it reduces exactly to the right-triangle definition because \(c\) is the hypotenuse and \(a\) is opposite \(\alpha\). 3. If \(\alpha=\beta\), then \(\sin(\alpha)=\sin(\beta)\). From \(\frac{a}{\sin(\alpha)}=\frac{b}{\sin(\beta)}\), it follows that \(a=b\).

1. \(\sin(\alpha)=\frac{a}{c}\) 2. The Law of Sines includes the right-triangle sine ratio as a special case, so there is no contradiction. 3. \(a=b\), which matches the equal-side property of an isosceles triangle.

- Use the Law of Sines to find a missing angle. - Check both an inverse-sine result and its supplement. - Test each candidate with the triangle angle sum.

1. For part a), the Law of Sines gives \(\sin(\gamma)=\frac{5.0\sin(100^\circ)}{8.0}\approx 0.6155\). 2. The possible values are \(\gamma_1\approx 37.99^\circ\) and \(\gamma_2\approx 142.01^\circ\). Since \(100^\circ+142.01^\circ>180^\circ\), the second value cannot form a triangle. Exactly one triangle is possible. 3. For part b), \(\sin(\beta)=\frac{6.0\sin(35^\circ)}{4.5}\approx 0.7648\). 4. The possible values are \(\beta_1\approx 49.89^\circ\) and \(\beta_2\approx 130.11^\circ\). Both satisfy \(35^\circ+\beta<180^\circ\), so two distinct triangles are possible.

a) Exactly one triangle is possible because the supplementary angle would make the angle sum exceed \(180^\circ\). b) Two distinct triangles are possible because both candidate values of \(\beta\) produce a valid third angle.

- Consider the shortest possible segment from the endpoint of side \(c\) to the ray forming the given angle. - The shortest distance from a point to a line is perpendicular. - What angle is formed when the minimum-length side is perpendicular to the other side?

1. The minimum possible value of \(a\) is the perpendicular distance from \(B\) to the ray containing side \(b\). This altitude is opposite \(\alpha\) in a right triangle with hypotenuse \(c\). 2. Therefore, \(a_{\min}=12.0\sin(40^\circ)\approx 7.71\,\text{cm}\). 3. At this minimum, side \(a\) is perpendicular to side \(b\), so the triangle is right and \(\gamma=90^\circ\).

1. The minimum length is approximately \(7.71\,\text{cm}\). 2. The triangle is a right triangle, and \(\gamma=90^\circ\).

- Use the Law of Sines to calculate \(\sin(\gamma)\). - A sine value can correspond to two supplementary angles between \(0^\circ\) and \(180^\circ\). - Check the triangle angle sum for both candidates. - Compare \(a\) with the altitude \(c\sin(\alpha)\) and with \(c\).

1. By the Law of Sines, \(\sin(\gamma)=\frac{7\sin(40^\circ)}{5}\approx 0.89990\). 2. The two possible angles are \(\gamma_1=\sin^{-1}(0.89990)\approx 64.15^\circ\) and \(\gamma_2=180^\circ-64.15^\circ\approx 115.85^\circ\). 3. Both are valid because \(40^\circ+64.15^\circ<180^\circ\) and \(40^\circ+115.85^\circ<180^\circ\). 4. Equivalently, the ambiguous-case condition \(c\sin(\alpha)<a<c\) holds: \(7\sin(40^\circ)<5<7\). Therefore, two noncongruent triangles exist.

a) \(\gamma\approx 64.15^\circ\) or \(\gamma\approx 115.85^\circ\) b) Both angles leave a positive third angle. Also, \(c\sin(\alpha)<a<c\), which is the two-triangle condition for this SSA case.

- Use the Law of Sines to find \(\sin(\beta)\). - Check both the inverse-sine result and its supplement. - Verify that each candidate leaves a positive third angle. - Compare \(c\) with \(b\sin(\gamma)\) and \(b\).

1. By the Law of Sines, \(\sin(\beta)=\frac{7.2\sin(38^\circ)}{5.4}\approx 0.82088\). 2. The two candidate angles are \(\beta_1\approx 55.17^\circ\) and \(\beta_2=180^\circ-55.17^\circ\approx 124.83^\circ\). 3. Both are valid because each gives \(\beta+38^\circ<180^\circ\). 4. The remaining angles are \(\alpha_1=180^\circ-38^\circ-55.17^\circ\approx 86.83^\circ\) and \(\alpha_2=180^\circ-38^\circ-124.83^\circ\approx 17.17^\circ\). 5. The full ambiguous-case condition also holds: \(b\sin(\gamma)<c<b\).

Two distinct triangles exist. Triangle 1: \(\alpha_1\approx 86.83^\circ\) and \(\beta_1\approx 55.17^\circ\). Triangle 2: \(\alpha_2\approx 17.17^\circ\) and \(\beta_2\approx 124.83^\circ\).

- Consider the two possible positions of the unknown vertex that are consistent with the given side and angle. - Use the Law of Sines to find all possible values of the angle opposite the \(120\,\text{m}\) side. - Check both candidate angles before finding the third side. - Compare the shorter given side with the relevant altitude and the longer side.

1. Let \(a=80\,\text{m}\), \(b=120\,\text{m}\), and \(\alpha=35^\circ\). 2. By the Law of Sines, \(\sin(\beta)=\frac{120\sin(35^\circ)}{80}\approx 0.86036\). 3. The possible angles are \(\beta_1\approx 59.36^\circ\) and \(\beta_2\approx 120.64^\circ\). Both produce a positive third angle because \(35^\circ+\beta<180^\circ\). 4. The corresponding third angles are \(\gamma_1\approx 85.64^\circ\) and \(\gamma_2\approx 24.36^\circ\). 5. Apply the Law of Sines again: \(c_1=\frac{80\sin(85.64^\circ)}{\sin(35^\circ)}\approx 139.07\,\text{m}\), and \(c_2=\frac{80\sin(24.36^\circ)}{\sin(35^\circ)}\approx 57.52\,\text{m}\). 6. The full ambiguity condition is \(b\sin(\alpha)<a<b\), which is satisfied here.

The third side is not unique because this SSA information satisfies \(b\sin(\alpha)<a<b\). The two possible lengths are approximately \(139.07\,\text{m}\) and \(57.52\,\text{m}\).

- In \(\triangle BCD\), use the Law of Sines and test the supplementary angle. - In \(\triangle ABD\), all three side lengths are known. - Does determining each triangle also determine which side of \(BD\) it occupies?

1. In \(\triangle BCD\), side \(BD = 6\,\text{cm}\) is opposite \(\angle C = 40^\circ\), and \(BC = 4\,\text{cm}\). 2. By the Law of Sines, \(\sin \angle D = \frac{4\sin 40^\circ}{6} \approx 0.4285\). This gives \(\angle D \approx 25.4^\circ\). The supplementary value is impossible because it would make the angle sum exceed \(180^\circ\). Thus, \(\triangle BCD\) is uniquely determined up to reflection. 3. Triangle \(ABD\) has side lengths \(5\,\text{cm}\), \(3\,\text{cm}\), and \(6\,\text{cm}\). These satisfy the triangle inequality, so \(\triangle ABD\) is uniquely determined by SSS, up to reflection. 4. The two triangles can be placed on the same side or on opposite sides of \(\overline{BD}\). Without an additional condition such as convexity, the entire quadrilateral is not uniquely determined.

Triangle \(BCD\) is uniquely determined by resolving the SSA case with the Law of Sines, and triangle \(ABD\) is uniquely determined by SSS. The quadrilateral is not unique because the component triangles can be placed on the same side or on opposite sides of \(BD\).