- Find the shared endpoint of the two labeled sides. - The side opposite an angle does not touch the vertex of that angle.

1. The included angle is at \(A\), where the labeled sides \(AB\) and \(AC\) meet. 2. The opposite side joins the other two vertices, so it is \(BC\).

\(BC\)

- Sketch both triangles with the \(50^\circ\) angle in the correct position. - Which sketch lets you determine the third side directly from the marked information?

1. The Law of Cosines finds a third side directly from two sides and their included angle. 2. Triangle X has this SAS information. In triangle Y, the given angle is not included, so these data do not fit the direct SAS form.

Triangle X. Its known angle is between the two known sides; triangle Y gives an angle opposite one of them.

- Substitute the cosine of a right angle into the formula. - Which term disappears?

1. Since \(\cos90^\circ=0\), the last term is zero. 2. The equation becomes \(a^2=b^2+c^2\), the Pythagorean theorem for a right triangle with hypotenuse \(a\).

The Pythagorean theorem: \(a^2=b^2+c^2\), because \(\cos90^\circ=0\).

- Recall the cosine value at \(90^\circ\). - What happens to a product when one factor is zero? - Which side is opposite the right angle?

1. \(\cos(90^\circ)=0\). 2. Substituting gives \(c^2=a^2+b^2-2ab(0)=a^2+b^2\), which is the Pythagorean theorem. 3. Since \(c\) is opposite \(\gamma\), when \(\gamma=90^\circ\), side \(c\) is the hypotenuse. The resulting equation is exactly the right-triangle relationship between the hypotenuse and legs.

1. \(\cos(90^\circ)=0\) 2. The Pythagorean theorem results: \(c^2=a^2+b^2\). 3. With \(\gamma=90^\circ\), \(c\) is the hypotenuse, so the formula matches right-triangle geometry.

- The given angle is between sides \(a\) and \(b\). - Use the Law of Cosines to find the side opposite \(\gamma\).

1. Use the Law of Cosines: \(c^2=a^2+b^2-2ab\cos(\gamma)\). 2. Substitute the given values: \(c^2=(8.5)^2+(12.0)^2-2\cdot 8.5\cdot 12.0\cos(42^\circ)\). 3. Thus, \(c=\sqrt{(8.5)^2+(12.0)^2-2\cdot 8.5\cdot 12.0\cos(42^\circ)}\approx 8.04\,\text{cm}\).

\(c\approx 8.04\,\text{cm}\)

- Read the three side lengths and the target angle from the diagram. - Use the Law of Cosines because all three side lengths are known. - Put the side opposite \(\gamma\) on the left side before solving for \(\cos(\gamma)\).

1. Use the Law of Cosines: \(c^2=a^2+b^2-2ab\cos(\gamma)\). 2. Substitute the side lengths from the diagram: \(7^2=5^2+8^2-2\cdot5\cdot8\cos(\gamma)\). 3. Simplify: \(49=89-80\cos(\gamma)\), so \(80\cos(\gamma)=40\). 4. Therefore, \(\cos(\gamma)=\frac{1}{2}=0.5\).

\(\cos(\gamma)=0.5\)

- The diagram gives two sides and their included angle. - Use the Law of Cosines to find the side opposite the marked angle. - Remember that cosine is negative for an obtuse angle.

1. Use the Law of Cosines with the two known sides and their included angle: \(x^2=3^2+5^2-2\cdot3\cdot5\cos(120^\circ)\). 2. Since \(\cos(120^\circ)=-\frac{1}{2}\), \(x^2=9+25+15=49\). 3. Therefore, \(x=7\,\text{cm}\).

\(7\,\text{cm}\)

- All three side lengths are shown in the diagram. - Solve the Law of Cosines for the cosine of the target angle. - A negative cosine is consistent with an obtuse angle.

1. By the Law of Cosines, \(c^2=a^2+b^2-2ab\cos(x)\). 2. Substitute the side lengths from the diagram: \(7^2=3^2+5^2-2\cdot3\cdot5\cos(x)\). 3. Thus, \(\cos(x)=\frac{3^2+5^2-7^2}{2\cdot3\cdot5}=-\frac{1}{2}\). 4. Therefore, \(x=\cos^{-1}\left(-\frac{1}{2}\right)=120^\circ\).

\(x=120^\circ\)

- What is the sign of cosine for an obtuse angle? - A minus sign in front of a negative value produces what sign? - Compare the resulting squares of the two positive lengths.

1. \(\cos120^\circ=-\frac{1}{2}\), so \(-2bc\cos120^\circ=bc>0\) for positive side lengths. 2. Therefore \(a^2=b^2+c^2+bc>b^2+c^2\), which means \(a>\sqrt{b^2+c^2}\).

The term is positive; \(a>\sqrt{b^2+c^2}\).

- Which angle is between the two given sides? - Substitute the exact cosine of that angle.

1. Let the opposite side be \(x\). Then \(x^2=3^2+8^2-2\cdot3\cdot8\cos60^\circ\). 2. Since \(\cos60^\circ=\frac{1}{2}\), \(x^2=9+64-24=49\). 3. Thus \(x=7\,\text{cm}\).

\(7\,\text{cm}\)

- Identify the angle between the two known sides. - What is the sign of the cosine of that angle? - The opposite side should be longer than either given side for this obtuse angle.

1. The two labeled sides meet at angle \(A=110^\circ\); the side opposite that angle is \(BC\). 2. \(BC^2=9^2+14^2-2\cdot9\cdot14\cos110^\circ\). 3. Therefore \(BC\approx19.1\,\text{cm}\).

\(BC\approx19.1\,\text{cm}\)

- Which side is opposite the angle you must find? - Rearrange the Law of Cosines to isolate the cosine of that angle. - Use inverse cosine after calculating the ratio.

1. The angle \(\theta\) lies opposite the \(9\,\text{cm}\) side. 2. \(9^2=6^2+7^2-2\cdot6\cdot7\cos\theta\). 3. \(\cos\theta=\frac{6^2+7^2-9^2}{2\cdot6\cdot7}=\frac{1}{21}\), so \(\theta\approx87.3^\circ\).

\(87.3^\circ\)

- The largest angle sits across from which side? - For that angle, what sign does the Law-of-Cosines cosine expression have?

1. The largest angle is opposite the longest side, \(10\). 2. The cosine of that angle has numerator \(5^2+7^2-10^2=25+49-100=-26\). 3. Thus \(\cos\theta=\frac{-26}{70}=-\frac{13}{35}<0\), making the largest angle obtuse.

\(\cos\theta=\frac{5^2+7^2-10^2}{2\cdot5\cdot7}=-\frac{13}{35}<0\). The largest angle is obtuse.

- Which law has a version that uses only three side lengths? - What known information must appear on both sides of a Law-of-Sines proportion?

1. With all three sides, the Law of Cosines can be rearranged to find the cosine of any angle. 2. The Law of Sines needs at least one known angle paired with its opposite side; no such angle is given. 3. Use the Law of Cosines first. After finding an angle, the Law of Sines can also be used for other angles.

Law of Cosines first. The Law of Sines needs a known opposite side-angle pair, which is not available yet.

- Compare the student's equation term by term with the Law of Cosines. - Keep the whole product involving the cosine together before evaluating.

1. The student's equation is missing the factor \(2\) in the cosine term. 2. The correct equation is \(x^2=5^2+7^2-2\cdot5\cdot7\cos60^\circ=74-35=39\). 3. Therefore \(x=\sqrt{39}\approx6.2\,\text{cm}\).

The factor \(2\) is missing. The third side is \(x\approx6.2\,\text{cm}\).

- Focus on the triangle formed by the named diagonal. - Find the angle between the two known sides of that triangle. - What sign does cosine have at an obtuse angle?

1. In triangle \(ABD\), the two known sides \(AB=10\,\text{cm}\) and \(AD=6\,\text{cm}\) meet at \(120^\circ\). 2. \(BD^2=10^2+6^2-2\cdot10\cdot6\cos120^\circ\). 3. Since \(\cos120^\circ=-\frac{1}{2}\), \(BD^2=100+36+60=196\), so \(BD=14\,\text{cm}\).

\(BD=14\,\text{cm}\)

- Which side is opposite the given angle? - In the Law of Cosines, what happens when the cosine of an obtuse angle is negative?

1. Let \(x\) be the side across from \(120^\circ\). Then \(x^2=4^2+7^2-2\cdot4\cdot7\cos120^\circ\). 2. Since \(\cos120^\circ=-\frac{1}{2}\), \(x^2=16+49+28=93\). 3. The length is \(x=\sqrt{93}\,\text{cm}\).

\(x^2=4^2+7^2-2\cdot4\cdot7\cos120^\circ=93\); \(x=\sqrt{93}\,\text{cm}\).

- Solve the Law of Cosines for the cosine of the angle opposite side \(c\). - Recall the sign of cosine for acute, right, and obtuse angles. - Compare \(a^2+b^2\) with \(c^2\).

1. Solve the Law of Cosines for \(\cos(\gamma)\): \(\cos(\gamma)=\frac{a^2+b^2-c^2}{2ab}\). 2. Substitute: \(\cos(\gamma)=\frac{6^2+9^2-13^2}{2\cdot6\cdot9}=\frac{36+81-169}{108}\). 3. Therefore, \(\cos(\gamma)=\frac{-52}{108}=-\frac{13}{27}\approx-0.4815\). 4. Since \(\cos(\gamma)<0\) for an angle between \(0^\circ\) and \(180^\circ\), \(\gamma\) is obtuse.

\(\cos(\gamma)=-\frac{13}{27}\approx-0.4815\). Because the cosine is negative, \(\gamma\) is obtuse.

- In an isosceles triangle, the two legs have equal lengths. - Substitute \(b=c=s\) into the Law of Cosines before calculating. - Pay attention to the sign of \(\cos(110^\circ)\).

1. Apply the Law of Cosines to the base: \(a^2=s^2+s^2-2s^2\cos(\alpha)\). 2. Factor the expression: \(a^2=2s^2(1-\cos(\alpha))\). 3. Substitute \(s=10\) and \(\alpha=110^\circ\): \(a=\sqrt{2(10)^2(1-\cos(110^\circ))}\approx16.38\,\text{cm}\). 4. In general, the simplified formula is \(a^2=2s^2(1-\cos(\alpha))\), or equivalently, \(a=s\sqrt{2(1-\cos(\alpha))}\).

a) \(a\approx16.38\,\text{cm}\) b) \(a^2=2s^2(1-\cos(\alpha))\), so \(a=s\sqrt{2(1-\cos(\alpha))}\).

- The largest angle is opposite the longest side. - Use the Law of Cosines because all three side lengths are known. - The sign of the cosine can help you classify the angle before finding its measure. - A triangle is obtuse when one interior angle is greater than \(90^\circ\).

1. The largest angle is \(\gamma\) because it is opposite the longest side, \(c=20\,\text{cm}\). 2. Solve the Law of Cosines for \(\cos(\gamma)\): \(\cos(\gamma)=\frac{a^2+b^2-c^2}{2ab}\). 3. Substitute: \(\cos(\gamma)=\frac{12^2+15^2-20^2}{2\cdot12\cdot15}=\frac{-31}{360}\). 4. Therefore, \(\gamma=\cos^{-1}\left(-\frac{31}{360}\right)\approx94.94^\circ\). 5. Since the largest angle is greater than \(90^\circ\), the triangle is obtuse.

The largest angle is \(\gamma\approx94.94^\circ\), so the triangle is obtuse.

- Read the three vertex coordinates from the graph. - Use the distance formula to find the three side lengths. - Apply the Law of Cosines to the side opposite angle \(\alpha\), then solve for the cosine.

1. Read the vertices as \(A(0, 0)\), \(B(5, 0)\), and \(C(3, 4)\). 2. The side lengths are \(AB=5\), \(AC=\sqrt{3^2+4^2}=5\), and \(BC=\sqrt{(3-5)^2+(4-0)^2}=\sqrt{20}\). 3. Apply the Law of Cosines with \(BC\) opposite \(\alpha\): \((\sqrt{20})^2=5^2+5^2-2\cdot5\cdot5\cos(\alpha)\). 4. Thus, \(20=50-50\cos(\alpha)\), so \(\cos(\alpha)=\frac{30}{50}=0.6\).

\(\cos(\alpha)=0.6\)

- The shorter diagonal and two adjacent sides form a triangle. - Read the two side lengths and their included angle from the diagram. - Use the Law of Cosines on that triangle.

1. The shorter diagonal forms a triangle with the two side lengths and the included \(60^\circ\) angle. 2. By the Law of Cosines, \(f^2=8^2+5^2-2\cdot8\cdot5\cos(60^\circ)\). 3. Since \(\cos(60^\circ)=\frac{1}{2}\), \(f^2=64+25-40=49\). 4. Therefore, \(f=7\,\text{cm}\).

\(7\,\text{cm}\)

- Identify the information pattern formed by the two known sides and the marked angle. - The unknown side lies opposite the marked included angle. - Substitute the three displayed values into the general-triangle relation that connects an SAS configuration to the opposite side.

1. The diagram gives two sides and their included angle, so apply the Law of Cosines to the side \(x\) opposite \(67^\circ\). 2. \(x^2=8^2+11^2-2\cdot8\cdot11\cos(67^\circ)\). 3. Therefore, \(x\approx10.7811\), so \(x\approx10.8\).

\(x\approx10.8\)

- The diagram gives SAS information, so first find the third side with the Law of Cosines. - Compare the resulting side length with the side labeled \(a\). - Equal sides in a triangle have equal opposite angles.

1. Use the Law of Cosines to find side \(b\): \(b^2=3^2+(3\sqrt{3})^2-2\cdot3\cdot3\sqrt{3}\cos(30^\circ)\). 2. Since \(\cos(30^\circ)=\frac{\sqrt{3}}{2}\), \(b^2=9+27-27=9\), so \(b=3\). 3. Because \(a=b\), the opposite angles are congruent. Therefore, \(x=30^\circ\).

\(x=30^\circ\)

- Inventory the given information before naming a method: note whether the triangle is right, whether any known side has its opposite angle, and whether two known sides meet at a known angle. - Eliminate any method that cannot use the displayed givens directly. - After choosing the method, write an equation that uses exactly the two known sides, the known angle, and the unknown side.

1. The diagram gives two sides and the included angle: \(AB=13\), \(AC=9\), and \(\angle A=58^\circ\). 2. This is an SAS configuration, so the Law of Cosines is the most direct method. 3. Let \(a=BC\). Then \(a^2=9^2+13^2-2\cdot9\cdot13\cos(58^\circ)\). 4. Therefore, \(a\approx11.2249\), so \(BC\approx11.2\) to the nearest tenth.

Law of Cosines: \(BC^2=9^2+13^2-2\cdot9\cdot13\cos(58^\circ)\), so \(BC\approx11.2\).

- Start with a relationship among the three sides and the included angle. - How would you solve an equation containing the unknown in more than one term? - Test any candidate length against the original triangle.

1. The Law of Cosines gives \(13^2=7^2+x^2-2\cdot7x\cos60^\circ\). 2. Simplifying gives \(169=49+x^2-7x\), or \(x^2-7x-120=0\). 3. Factor: \((x-15)(x+8)=0\), so \(x=15\) or \(x=-8\). 4. Side lengths must be positive, so only \(x=15\,\text{cm}\) is valid.

\(x=15\,\text{cm}\)

- The diagonal divides the quadrilateral into two triangles whose three side lengths are shown. - Apply the Law of Cosines separately in the two triangles. - At \(B\) and \(D\), add the two smaller triangle angles that meet at the same vertex.

1. In \(\triangle ABD\), the Law of Cosines gives \(\cos(\alpha)=\frac{15^2+13^2-14^2}{2\cdot15\cdot13}\), so \(\alpha\approx59.5^\circ\). 2. In \(\triangle BCD\), \(\cos(\gamma)=\frac{12^2+10^2-14^2}{2\cdot12\cdot10}\), so \(\gamma\approx78.5^\circ\). 3. At \(B\), \(\angle ABD\approx53.1^\circ\) and \(\angle DBC\approx44.4^\circ\). Therefore, \(\beta\approx97.5^\circ\). 4. At \(D\), the two triangle angles are approximately \(67.4^\circ\) and \(57.1^\circ\). Therefore, \(\delta\approx124.5^\circ\). 5. The four angles total \(360.0^\circ\), which checks the result.

\(\alpha\approx59.5^\circ\) \(\beta\approx97.5^\circ\) \(\gamma\approx78.5^\circ\) \(\delta\approx124.5^\circ\)

- The diagram gives two sides with an included angle and the side opposite that angle. - Applying the Law of Cosines produces a quadratic equation in \(x\). - Use the condition \(x>4\) only after solving the quadratic.

1. Apply the Law of Cosines to the side labeled \(7\): \(7^2=x^2+8^2-2\cdot x\cdot8\cos(60^\circ)\). 2. Simplify: \(49=x^2+64-8x\), so \(x^2-8x+15=0\). 3. Factor: \((x-3)(x-5)=0\), giving \(x=3\) or \(x=5\). 4. Because \(x>4\), the required length is \(x=5\).

\(x=5\)

- The given information is SAS, so first find the third side. - After that, a side-angle opposite pair is available for the Law of Sines. - In the final inverse-sine step, compare side lengths to decide which angle is possible.

1. First use the Law of Cosines to find \(c\): \(c^2=2^2+(1+\sqrt{3})^2-2\cdot2(1+\sqrt{3})\cos(60^\circ)=6\). Thus, \(c=\sqrt{6}\). 2. By the Law of Sines, \(\frac{\sin(x)}{2}=\frac{\sin(60^\circ)}{\sqrt{6}}\), so \(\sin(x)=\frac{1}{\sqrt{2}}\). 3. This gives \(x=45^\circ\) or \(135^\circ\). Since the side opposite \(x\) is shorter than \(c=\sqrt{6}\), the opposite angles satisfy \(x<60^\circ\). 4. Therefore, \(x=45^\circ\).

\(x=45^\circ\)

- Work with the acute and obtuse configurations separately at first. - In the obtuse panel, consider the acute angle between the altitude's base extension and side \(AC\). - Express the horizontal distance from the altitude foot to \(B\) using cosine. - After using Pythagoras, look for a familiar identity involving sine and cosine squares.

1. In panel a), \(AD=b\cos A\) and \(CD=b\sin A\), so \(BD=c-b\cos A\). 2. Apply Pythagoras to triangle \(BCD\): \(a^2=(c-b\cos A)^2+(b\sin A)^2\). 3. Expand and use \(\sin^2 A+\cos^2 A=1\) to get \(a^2=b^2+c^2-2bc\cos A\). 4. In panel b), let \(E\) be the exterior altitude foot. The acute angle at \(A\) in triangle \(ACE\) is \(180^\circ-A\). 5. Thus, \(AE=b\cos(180^\circ-A)=-b\cos A\) and \(CE=b\sin(180^\circ-A)=b\sin A\). 6. Since \(BE=BA+AE=c-b\cos A\), Pythagoras in triangle \(BCE\) gives the same squared expression and the same Law of Cosines. 7. At \(A=90^\circ\), \(\cos A=0\), leaving \(a^2=b^2+c^2\), the Pythagorean theorem.

a) \(BD=c-b\cos A\) and \(CD=b\sin A\), so \(a^2=b^2+c^2-2bc\cos A\). b) \(AE=-b\cos A\), hence \(BE=c-b\cos A\) and \(CE=b\sin A\), producing the same formula. c) If \(A=90^\circ\), then \(\cos A=0\) and \(a^2=b^2+c^2\).