1. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines 8-5 Law of Sines and Law of Cosines Holt Geometry Warm Up Warm Up Lesson Presentation Lesson Presentation Lesson Quiz Lesson Quiz Holt McDougal Geometry

2. 8-5 Law of Sines and Law of Cosines Warm Up 1. What is the third angle measure in a triangle with angles measuring 65° and 43°? Find each value. Round trigonometric ratios to the nearest hundredth and angle measures to the nearest degree. 2. sin 73°3. cos 18°4. tan 82° 5. sin -1 (0.34)6. cos -1 (0.63)7. tan -1 (2.75) 72° 0.960.957.12 20°51°70°

3. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines Use the Law of Sines and the Law of Cosines to solve triangles. Objective

4. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines 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.

5. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines 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°  –4.33cos 165°  –0.97sin 93°  1.00

6. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines Check It Out! Example 1 Use a calculator to find each trigonometric ratio. Round to the nearest hundredth. a. tan 175° tan 175°  –0.09 b. cos 92°c. sin 160° cos 92°  –0.03sin 160°  0.34

7. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines 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).

8. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines 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. Cross Products Property Divide both sides by sin 39. FG sin 39° = 40 sin 32°

9. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines Example 2B: 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 Substitute the given values. Multiply both sides by 6. Use the inverse sine function to find mQ.

10. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines 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. Cross Products Property Divide both sides by sin 39°. NP sin 39° = 22 sin 88°

11. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines Check It Out! Example 2b Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mLmL Law of Sines Substitute the given values. Cross Products Property Use the inverse sine function to find mL. 10 sin L = 6 sin 125°

12. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines Check It Out! Example 2c Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mXmX Law of Sines Substitute the given values. Cross Products Property Use the inverse sine function to find mX. 7.6 sin X = 4.3 sin 50°

13. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines 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° mA + 67° + 44° = 180° mA = 69° Prop of ∆. Substitute the given values. Simplify.

14. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines Check It Out! Example 2D Continued Law of Sines Substitute the given values. Cross Products Property 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°

15. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines 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.

16. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines You can use the Law of Cosines to solve a triangle if you are given two side lengths and the included angle measure (SAS) or three side lengths (SSS).

17. Holt McDougal Geometry 8-5 Law of Sines and 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 XZ 2 = XY 2 + YZ 2 – 2(XY)(YZ)cos Y = 35 2 + 30 2 – 2(35)(30)cos 110° XZ 2  2843.2423 XZ  53.3 Law of Cosines Substitute the given values. Simplify. Find the square root of both sides.

18. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines Example 3B: 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 Law of Cosines Substitute the given values. Simplify. Subtract 290 both sides.

19. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines Example 3B Continued Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mTmT –241 = –286 cosT Solve for cosT. Use the inverse cosine function to find mT.

20. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines Check It Out! Example 3a 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 Substitute the given values. Simplify. Find the square root of both sides.

21. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines Check It Out! Example 3b 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. Simplify. Subtract 325 both sides.

22. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines Check It Out! Example 3b 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.

23. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines Check It Out! Example 3c Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. YZ YZ 2 = XY 2 + XZ 2 – 2(XY)(XZ)cos X = 10 2 + 4 2 – 2(10)(4)cos 34° YZ 2  49.6770 YZ  7.0 Law of Cosines Substitute the given values. Simplify. Find the square root of both sides.

24. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines 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

25. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines Example 4: Sailing Application A sailing club has planned a triangular racecourse, as shown in the diagram. How long is the leg of the race along BC? How many degrees must competitors turn at point C? Round the length to the nearest tenth and the angle measure to the nearest degree.

26. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines Example 4 Continued Step 1 Find BC. BC 2 = AB 2 + AC 2 – 2(AB)(AC)cos A = 3.9 2 + 3.1 2 – 2(3.9)(3.1)cos 45° BC 2  7.7222 BC  2.8 mi Law of Cosines Substitute the given values. Simplify. Find the square root of both sides.

27. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines Step 2 Find the measure of the angle through which competitors must turn. This is mC. Example 4 Continued Law of Sines Substitute the given values. Multiply both sides by 3.9. Use the inverse sine function to find mC.

28. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines Check It Out! Example 4 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

29. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines Check It Out! Example 4 Continued Step 1 Find the length of the cable. AC 2 = AB 2 + BC 2 – 2(AB)(BC)cos B = 31 2 + 40 2 – 2(31)(40)cos 100° AC 2  2991.6474..... AC 54.7 m Law of Cosines Substitute the given values. Simplify. Find the square root of both sides.

30. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines Step 2 Find the measure of the angle the cable would make with the ground. Check It Out! Example 4 Continued Law of Sines Substitute the given values. Multiply both sides by 56. Use the inverse sine function to find mA.

31. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines 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

32. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines 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. 6. a = 20, b = 15, and c = 8.3. Find mA. Lesson Quiz: Part II 477.2 21.6 115°

33. Holt McDougal Geometry 8-5 Law of Sines and Law of Cosines Lesson Quiz: Part III 7. 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°