1. Splash Screen

2. Lesson Menu Five-Minute Check (over Lesson 8–5) NGSSS Then/Now New Vocabulary Theorem 8.10: Law of Sines Example 1: Law of Sines (AAS or ASA) Example 2: Law of Sines (SSA) Theorem 8.11: Law of Cosines Example 3: Law of Cosines (SAS) Example 4: Law of Cosines (SSS) Example 5: Real-World Example: Indirect Measurement Example 6: Solve a Triangle Concept Summary: Solving a Triangle

3. Over Lesson 8–5 A.A B.B C.C D.D 5-Minute Check 1 A.  URT B.  SRT C.  RST D.  SRU Name the angle of depression in the figure.

4. Over Lesson 8–5 A.A B.B C.C D.D 5-Minute Check 2 A.about 70.6° B.about 60.4° C.about 29.6° D.about 19.4° Find the angle of elevation of the Sun when a 6-meter flagpole casts a 17-meter shadow.

5. Over Lesson 8–5 A.A B.B C.C D.D 5-Minute Check 3 A.about 1.8° B.about 2.4° C.about 82.4° D.about 88.6° After flying at an altitude of 575 meters, a helicopter starts to descend when its ground distance from the landing pad is 13.5 kilometers. What is the angle of depression for this part of the flight?

6. Over Lesson 8–5 A.A B.B C.C D.D 5-Minute Check 4 A.about 81.4 ft B.about 236.4 ft C.about 726 ft D.about 804 ft The top of a signal tower is 250 feet above sea level. The angle of depression from the top of the tower to a passing ship is 19°. How far is the foot of the tower from the ship?

7. Over Lesson 8–5 A.A B.B C.C D.D 5-Minute Check 5 A.50 ft B.104 ft C.1060 ft D.4365 ft Jay is standing 50 feet away from the Eiffel Tower and measures the angle of elevation to the top of the tower as 87.3°. Approximately how tall is the Eiffel Tower?

8. NGSSS MA.912.T.2.1 Define and use the trigonometric ratios (sine, cosine, tangent, cotangent, secant, cosecant) in terms of angles of right triangles.

9. Then/Now You used trigonometric ratios to solve right triangles. (Lesson 8–4) Use the Law of Sines to solve triangles. Use the Law of Cosines to solve triangles.

10. Vocabulary Law of Sines Law of Cosines

11. Concept

12. Example 1 Law of Sines (AAS or ASA) Find p. Round to the nearest tenth. We are given measures of two angles and a nonincluded side, so use the Law of Sines to write a proportion.

13. Example 1 Law of Sines (AAS or ASA) Law of Sines Use a calculator. Divide each side by sin Cross Products Property Answer: p ≈ 4.8

14. A.A B.B C.C D.D Example 1 A.4.6 B.29.9 C.7.8 D.8.5 Find c to the nearest tenth.

15. Example 2 Law of Sines (SSA) Find x. Round to the nearest degree.

16. Example 2 Law of Sines (SSA) Law of Sines m  B = 50, b = 10, a = 11 Cross Products Property Divide each side by 10. Use a calculator. Use the inverse sine ratio. Answer: x ≈ 57.4

17. A.A B.B C.C D.D Example 2 A.39 B.43 C.46 D.49 Find x. Round to the nearest degree.

18. Concept

19. Example 3 Law of Cosines (SAS) Find x. Round to the nearest tenth. Use the Law of Cosines since the measures of two sides and the included angle are known.

20. Example 3 Law of Cosines (SAS) Answer: x ≈ 18.9 Simplify. Take the square root of each side. Law of Cosines Use a calculator.

21. A.A B.B C.C D.D Example 3 A.25.1 B.44.5 C.22.7 D.21.1 Find r if s = 15, t = 32, and m  R = 40. Round to the nearest tenth.

22. Example 4 Law of Cosines (SSS) Find m  L. Round to the nearest degree. Law of Cosines Simplify.

23. Example 4 Law of Cosines (SSS) Answer: m  L ≈ 49 Solve for L. Use a calculator. Subtract 754 from each side. Divide each side by –270.

24. A.A B.B C.C D.D Example 4 A.44° B.51° C.56° D.69° Find m  P. Round to the nearest degree.

25. Example 5 Indirect Measurement AIRCRAFT From the diagram of the plane shown, determine the approximate width of each wing. Round to the nearest tenth meter.

26. Example 5 Indirect Measurement Cross products Law of Sines Use the Law of Sines to find KJ.

27. Example 5 Indirect Measurement Answer: The width of each wing is about 16.9 meters. Simplify. Divide each side by sin.

28. A.A B.B C.C D.D Example 5 A.93.5 in. B.103.5 in. C.96.7 in. D.88.8 in. The rear side window of a station wagon has the shape shown in the figure. Find the perimeter of the window if the length of DB is 31 inches. Round to the nearest tenth.

29. Example 6 Solve a Triangle Solve triangle PQR. Round to the nearest degree. Since the measures of three sides are given (SSS), use the Law of Cosines to find m  P. p 2 =r 2 + q 2 – 2pq cos PLaw of Cosines 8 2 =9 2 + 7 2 – 2(9)(7) cos Pp = 8, r = 9, and q = 7

30. Example 6 Solve a Triangle 64=130 – 126 cos PSimplify. –66=–126 cos PSubtract 130 from each side. Divide each side by –126. Use the inverse cosine ratio. Use a calculator.

31. Example 6 Solve a Triangle Use the Law of Sines to find m  Q. Law of Sines Multiply each side by 7. Use the inverse sine ratio. Use a calculator. m  P ≈ 58, p = 8, q = 7.

32. Example 6 Solve a Triangle Answer: Therefore, m  P ≈ 58; m  Q ≈ 48 and m  R ≈ 74. By the Triangle Angle Sum Theorem, m  R ≈ 180 – (58 + 48) or 74.

33. A.A B.B C.C D.D Example 6 A.m  R = 82, m  S = 58, m  T = 40 B.m  R = 58, m  S = 82, m  T = 40 C.m  R = 82, m  S = 40, m  T = 58 D.m  R = 40, m  S = 58, m  T = 82 Solve ΔRST. Round to the nearest degree.

34. Concept

35. End of the Lesson