1. Lesson 8.5 1 The Law of Sines and the Law of Cosines Use the Law of Sines to solve triangles. Objective

2. Holt McDougal Geometry 2 Ex1: 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°  –4.33cos 165°  –0.97sin 93°  1.00 In this lesson, you will learn to solve any triangle. To do so, you will need to calculate trigonometric ratios for angle measures up to 180°. You can use a calculator to find these values.

3. Holt McDougal Geometry 3 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).

4. Holt McDougal Geometry 4 Example 2: 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 FG sin 39° = 40 sin 32°

5. Holt McDougal Geometry 5 Check It Out! Example 3 Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. AC mA + mB + mC = 180° mA + 67° + 44° = 180° mA = 69° Prop of ∆.

6. Holt McDougal Geometry 6 Check It Out! Example 3 Continued Law of Sines Divide both sides by sin 69°. Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. AC sin 69° = 18 sin 67° 69°

7. Holt McDougal Geometry 7 Check It Out! Example 4 Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mXmX Law of Sines Use the inverse sine function to find mX. 7.6 sin X = 4.3 sin 50°

8. Holt McDougal Geometry 8 Example 5: Using the Law of Sines Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mQmQ Law of Sines

9. Holt McDougal Geometry 9 Lesson Quiz: Part I Use a calculator to find each trigonometric ratio. Round to the nearest hundredth. 1. tan 154° 2. cos 124° 3. sin 162° –0.49 –0.56 0.31

10. Holt McDougal Geometry 10 Use ΔABC for Items 4–6. Round lengths to the nearest tenth and angle measures to the nearest degree. 4. mB = 20°, mC = 31° and b = 210. Find a. 5. a = 16, b = 10, and mC = 110°. Find c. Lesson Quiz 477.2 21.6

11. Law of Sines 8.5 Day Two 11 Use the Law of Cosines to solve triangles. Objective

12. Holt McDougal Geometry 12 The Law of Sines cannot be used to solve every ∆. Instead, you can apply the Law of Cosines. You can use the Law of Cosines to solve a ∆ if you are given: two side lengths and the included angle measure (SAS) or three side lengths (SSS).

13. Holt McDougal Geometry 13 Example 1: Using the Law of Cosines Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. XZ XZ 2 = XY 2 + YZ 2 – 2(XY)(YZ)cos Y XZ 2 = 35 2 + 30 2 – 2(35)(30)cos 110° XZ 2  2843.2423 XZ  53.3 Law of Cosines Substitute

14. Holt McDougal Geometry 14 Example 2 Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. DE DE 2 = EF 2 + DF 2 – 2(EF)(DF)cos F = 18 2 + 16 2 – 2(18)(16)cos 21° DE 2  42.2577 DE  6.5 Law of Cosines

15. Holt McDougal Geometry 15 Example 3: Using the Law of Cosines Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mTmT RS 2 = RT 2 + ST 2 – 2(RT)(ST)cos T 7 2 = 13 2 + 11 2 – 2(13)(11)cos T 49 = 290 – 286 cosT –241 = –286 cosT

16. Holt McDougal Geometry 16 Example 4 Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mKmK JL 2 = LK 2 + KJ 2 – 2(LK)(KJ)cos K 8 2 = 15 2 + 10 2 – 2(15)(10)cos K 64 = 325 – 300 cosK –261 = –300 cosK Law of Cosines Substitute the given values.

17. Holt McDougal Geometry 17 Example 4 Continued Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mKmK –261 = –300 cosK Solve for cosK. Use the inverse cosine function to find mK.

18. Holt McDougal Geometry 18 Do not round your answer until the final step of the computation. If a problem has multiple steps, store the calculated answers to each part in your calculator. Helpful Hint

19. Holt McDougal Geometry 19 Example 5 What if…? Another engineer suggested using a cable attached from the top of the tower to a point 31 m from the base. How long would this cable be, and what angle would it make with the ground? Round the length to the nearest tenth and the angle measure to the nearest degree. 31 m Step 1 Find the length of the cable. AC 2 = AB 2 + BC 2 – 2(AB)(BC)cos B AC 2 = 31 2 + 56 2 – 2(31)(56)cos 100° AC 2  4699.9065 AC 68.6 m

20. Holt McDougal Geometry 20 Step 2 Find the measure of the angle the cable would make with the ground. Example 5 Continued Law of Sines 31 m

21. Holt McDougal Geometry 21 Use ΔABC for Items 4–6. Round lengths to the nearest tenth and angle measures to the nearest degree. 1. a = 20, b = 15, and c = 8.3. Find mA. Lesson Quiz 115°

22. Holt McDougal Geometry 22 Lesson Quiz 2. An observer in tower A sees a fire 1554 ft away at an angle of depression of 28°. To the nearest foot, how far is the fire from an observer in tower B? To the nearest degree, what is the angle of depression to the fire from tower B? 1212 ft; 37°