Objective Use the Law of Sines and the Law of Cosines to solve triangles.

Example 1: Finding Trigonometric Ratios for Obtuse Angles Use your calculator to find each trigonometric ratio. Round to the nearest hundredth. A. tan 103° B. cos 165° C. sin 93° tan 103°  –4.33 cos 165°  –0.97 sin 93°  1.00

Check It Out! Example 1 Use a calculator to find each trigonometric ratio. Round to the nearest hundredth. a. tan 175° b. cos 92° c. sin 160° tan 175°  –0.09 cos 92°  –0.03 sin 160°  0.34

You can use the Law of Sines to solve a triangle if you are given • two angle measures and any side length (ASA or AAS) or • two side lengths and a non-included angle measure (SSA).

Example 2A: Using the Law of Sines Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. FG Law of Sines Substitute the given values. FG sin 39° = 40 sin 32° Cross Products Property Divide both sides by sin 39.

Example 2B: Using the Law of Sines Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mQ Law of Sines Substitute the given values. Multiply both sides by 6. Use the inverse sine function to find mQ.

Check It Out! Example 2a Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. NP Law of Sines Substitute the given values. NP sin 39° = 22 sin 88° Cross Products Property Divide both sides by sin 39°.

Check It Out! Example 2b Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mL Law of Sines Substitute the given values. Cross Products Property 10 sin L = 6 sin 125° Use the inverse sine function to find mL.

Check It Out! Example 2c Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mX Law of Sines Substitute the given values. 7.6 sin X = 4.3 sin 50° Cross Products Property Use the inverse sine function to find mX.

Check It Out! Example 2d Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. AC mA + mB + mC = 180° Prop of ∆. mA + 67° + 44° = 180° Substitute the given values. mA = 69° Simplify.

Check It Out! Example 2D Continued Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. Law of Sines Substitute the given values. AC sin 69° = 18 sin 67° Cross Products Property Divide both sides by sin 69°.

The Law of Sines cannot be used to solve every triangle The Law of Sines cannot be used to solve every triangle. If you know two side lengths and the included angle measure or if you know all three side lengths, you cannot use the Law of Sines. Instead, you can apply the Law of Cosines.

Example 3A: Using the Law of Cosines Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. XZ XZ2 = XY2 + YZ2 – 2(XY)(YZ)cos Y Law of Cosines Substitute the given values. = 352 + 302 – 2(35)(30)cos 110° XZ2  2843.2423 Simplify. Find the square root of both sides. XZ  53.3

Check It Out! Example 3a Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. DE DE2 = EF2 + DF2 – 2(EF)(DF)cos F Law of Cosines Substitute the given values. = 182 + 162 – 2(18)(16)cos 21° DE2  42.2577 Simplify. Find the square root of both sides. DE  6.5

Example 3B: Using the Law of Cosines Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mT RS2 = RT2 + ST2 – 2(RT)(ST)cos T Law of Cosines Substitute the given values. 72 = 132 + 112 – 2(13)(11)cos T 49 = 290 – 286 cosT Simplify. Subtract 290 both sides. –241 = –286 cosT

Example 3B Continued Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mT –241 = –286 cosT Solve for cosT. Use the inverse cosine function to find mT.

Check It Out! Example 3b Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mK JL2 = LK2 + KJ2 – 2(LK)(KJ)cos K Law of Cosines Substitute the given values. 82 = 152 + 102 – 2(15)(10)cos K 64 = 325 – 300 cosK Simplify. Subtract 325 both sides. –261 = –300 cosK

Check It Out! Example 3b Continued Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mK –261 = –300 cosK Solve for cosK. Use the inverse cosine function to find mK.

Check It Out! Example 3c Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. YZ YZ2 = XY2 + XZ2 – 2(XY)(XZ)cos X Law of Cosines Substitute the given values. = 102 + 42 – 2(10)(4)cos 34° YZ2  49.6770 Simplify. Find the square root of both sides. YZ  7.0

Check It Out! Example 3d Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mR PQ2 = PR2 + RQ2 – 2(PR)(RQ)cos R Law of Cosines Substitute the given values. 9.62 = 5.92 + 10.52 – 2(5.9)(10.5)cos R 92.16 = 145.06 – 123.9cosR Simplify. Subtract 145.06 both sides. –52.9 = –123.9 cosR

Check It Out! Example 3d Continued Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mR –52.9 = –123.9 cosR Solve for cosR. Use the inverse cosine function to find mR.