1. Module 8 Lesson 5 Oblique Triangles Florben G. Mendoza

2. If none of the angles of a triangle is a right angle, the triangle is called oblique. All angles are acute Two acute angles, one obtuse angle Florben G. Mendoza

3. To solve an oblique triangle means to find the lengths of its sides and the measurements of its angles. ab c A B C Sides:a b c Angles: A B C Florben G. Mendoza

4. FOUR CASES CASE 1: One side and two angles are known (SAA or ASA). CASE 2: Two sides and the angle opposite one of them are known (SSA). CASE 3: Two sides and the included angle are known (SAS). CASE 4: Three sides are known (SSS). Florben G. Mendoza

5. CASE 1: ASA or SAA S A A ASA S A A SAA Florben G. Mendoza

6. S S A CASE 2: SSA Florben G. Mendoza

7. S S A CASE 3: SAS Florben G. Mendoza

8. S S S CASE 4: SSS Florben G. Mendoza

9. 9 Practice Exercise 1: 1) 2) 3) 4) 5) 6) 7) 8) 9) SAS SSA ASA SAA SSS ASA Florben G. Mendoza

10. 10 Practice Exercise 2: 1. (A, B, c) 2. (A, B, a) 3. (b, c, A) 4. (a, b, A) 5. (a, b, c) 6. (C, b, c) 7. (a, B, C) 8. (a, A, C) 9. (A, b, C) 10. (C, b, a) SAA ASA SAS SSA SSS SSA ASA SAA ASA SAS Florben G. Mendoza

11. The Law of Sines is used to solve triangles in which Case 1 or 2 holds. That is, the Law of Sines is used to solve SAA, ASA or SSA triangles. ASA A A S SAA S A A SSA S A S Florben G. Mendoza

12. Law of Sines A B C a b c Let’s drop an altitude and call it h. h If we think of h as being opposite to both A and B, then Let’s solve both for h. This means Florben G. Mendoza

13. A B C a b c If I were to drop an altitude to side a, I could come up with Putting it all together gives us the Law of Sines. You can also use it upside-down. Florben G. Mendoza

14. Example 1: A B C a b c 45° 50° = 30 = 180° - (45° + 50°) Step 1: C = 180° - (A + B) C = 85° = 180° - 95° Step 2: a sin A = b sin B 30 sin 45° = b sin 50° b (sin 45°) = 30 (sin 50°) sin 45° b = 32.50 SAA Florben G. Mendoza

15. Example 1: SAA A B C a b c 45° 50° = 30 Step 3: a sin A = c sin C 30 sin 45° = c sin 85° c (sin 45°) = 30 (sin 85°) sin 45° c = 42.26 Florben G. Mendoza

16. Example 2: Let C = 35°, B = 10°, and a = 45 Step 1: A = 180° - (B + C) = 180° - (10° + 35°) = 180° - 45° A = 135° A B a b c 35° 10° = 45 C Step 2: a sin A = b sin B 45 sin 135° = b sin 10° b (sin 135°) = 45 (sin 10°) sin 135° b = 11.05 ASA Florben G. Mendoza

17. Example 2: ASA Let C = 35°, B = 10°, and a = 45 A B a b c 35° 10° = 45 C Step 3: a sin A = c sin C 45 sin 135° = c sin 35° c (sin 135°) = 45 (sin 35°) sin 135° c = 36.50 Florben G. Mendoza

18. Ambiguous Case (SSA) Case 1: If A is acute and a < b A C B b a c h = b sin A a. If a < b sinA A C B b a c h NO SOLUTION Florben G. Mendoza

19. Case 1: If A is acute and a < b AC B b a c h = b sin A b. If a = b sinA A C B b = a c h 1 SOLUTION Ambiguous Case (SSA) Florben G. Mendoza

20. Case 1: If A is acute and a < b AC B ba c h = b sin A b. If a > b sinA A C B b c h 2 SOLUTIONS aa B   180 -  Ambiguous Case (SSA) Florben G. Mendoza

21. Case 2: If A is obtuse and a > b C A B a b c ONE SOLUTION Ambiguous Case (SSA) Florben G. Mendoza

22. Case 2: If A is obtuse and a ≤ b C A B a b c NO SOLUTION Ambiguous Case (SSA) Florben G. Mendoza

23. Let A = 40°, b = 10, and a = 9 Example 3: A B C a b c h = 10 = 9 40° Step 1: Solve for h h = b sin A h = 10 sin 40° h = 6.43 a > h ( 2 Solutions) Step 2: a sin A = b sin B 9 sin 40° = 10 sin B 9 (sin B) = 10 (sin 40°) 9 9 sin B = 0.71 B = sin -1 0.71 B = 45.23° SSA Florben G. Mendoza

24. Let A = 40°, b = 10, and a = 9 Example 3: SSA A B C a b c h = 10 = 9 40° Step 3: C = 180° - (A + B) C = 180° - (40° + 45.23°) C = 180° - 85.23° C = 94.77° a sin A = c sin C 9 sin 40° = c sin 94.77° c (sin 40°) = 9 (sin 94.77°) c = 13.95 Step 4: (sin 40°) Florben G. Mendoza

25. Let A = 40°, b = 10, and a = 9 Example 3: SSA A B C a b c h = 10 = 9 40° b = 10 a = 9 45.23° B A C 9 Step 5: B’ = 180° - B B’ = 134.77° Step 6: C’ = 180° - (A + B’) B’ = 180° - 45.23° C’ = 180° - (40° + 134.77°) C’ = 180° - 174.77° C’ = 5.23° A B’ C’ c’ 40° 9 b = 10 2 ND Solution Florben G. Mendoza

26. Let A = 40°, b = 10, and a = 9 Example 3: SSA Step 7: a sin A = c’ sin C’ 9 sin 40° = c’ sin 5.23° c’ (sin 40°) = 9 (sin5.23°) (sin 40°) c’ = 1.28 (sin 40°) A B’ C’ c’ 40° 9 b = 10 Florben G. Mendoza

27. Let B = 53°, b = 10, and c = 32 Example 4: SSA Step 1: Solve for h h = c sin B h = 32 sin 53° h = 40.07 b < h ( No Solution) A C B b a c h Florben G. Mendoza

28. 28 Example 5: SSA Let C = 100°, a = 25, and c = 33 Step 1: c sin C = a sin A 33 sin 100° = 25 sin A 33(sin A ) = 25 (sin 100°) 33 sin A = 0.75 A = sin -1 0.75 A = 48.59° Step 2: B = 180° - (A + C) B = 180° - (48.59° + 100°) B = 180° - 148.59° B = 31.41° C A B 100° 25 33 b Florben G. Mendoza

29. 29 Example 5: SSA Let C = 100°, a = 25, and c = 33 C A B 100° 25 33 Step 3: c sin C = b sin B 33 sin 100° = b sin 31.41° 33(sin 31.41° ) = b(sin 100°) sin 100° b = 17.46 Florben G. Mendoza

30. 30 Example 6: SSA Let A = 133°, a = 27, and c = 40 A B C 133° 27 40 a < c (No Solution) Florben G. Mendoza

31. We use the Law of Sines to solve CASE 1 (SAA or ASA) and CASE 2 (SSA) of an oblique triangle. The Law of Cosines is used to solve CASES 3 and 4. CASE 3: Two sides and the included angle are known (SAS). CASE 4: Three sides are known (SSS). Florben G. Mendoza

32. 32 Deriving the Law of Cosines Write an equation using Pythagorean theorem for shaded triangle. b h a k c - k A B C c Florben G. Mendoza

33. 33 Law of Cosines Similarly Note the pattern Florben G. Mendoza

34. 34 Law of Cosines a 2 = b 2 + c 2 – 2bc cos A 2bc cos A = b 2 + c 2 – a 2 2bc cos A = b 2 + c 2 - a 2 2bc Similarly; cos A = b 2 + c 2 - a 2 2bc cos B = a 2 + c 2 - b 2 2ac cos C = a 2 + b 2 - c 2 2ab Florben G. Mendoza

35. 35 Example 7: SAS Let A = 42°, b = 12.9 & c = 15.4 Step 1: a 2 = b 2 + c 2 – 2bc cos A a 2 = (12.9) 2 + (15.4) 2 – 2 (12.9) (15.4) (cos 42°) a 2 = 403.57 – 295.27 a 2 = 108.3 a =10.41 A B C 42° 15.4 12.9 a Florben G. Mendoza

36. 36 Step 2: cos B = a 2 + c 2 - b 2 2ac cos B = (10.41) 2 + (15.4) 2 – (12.9) 2 2(10.41)(15.4) Example 3: Let A = 42°, b = 12.9 & c = 15.4 A B C 42° 15.4 12.9 a cos B = 179.12 320.68 cos B = 0.56 B = cos -1 0.56 B = 55.94° SAS Florben G. Mendoza

37. 37 Example 3: Let A = 42°, b = 12.9 & c = 15.4 A B C 42° 15.4 12.9 a SAS Step 3: C = 180° - (A + B) C = 180° - (42° + 55.94°) C = 180° - 97.94° C = 82.06° Florben G. Mendoza

38. 38 Example 8: SSS Let a = 9.47, b = 15.9 & c = 21.1 Step 1: cos A = b 2 + c 2 - a 2 2bc cos A = (15.9) 2 + (21.1) 2 – (9.47) 2 2(15.9)(21.1) cos A = 608.34 670.98 cos A = 0.91 A = cos -1 0.91 A = 24.49° Florben G. Mendoza

39. 39 Step 2: cos B = a 2 + c 2 - b 2 2ac Example 8: SSS Let a = 9.47, b = 15.9 & c = 21.1 cos B = (9.47) 2 + (21.1) 2 – (15.9) 2 2(9.47)(21.1) cos B = 282.08 399.63 cos B = 0.71 B = cos -1 0.71 B = 44.77° Florben G. Mendoza

40. 40 Example 8: SSS Let a = 9.47, b = 15.9 & c = 21.1 Step 3: C = 180° - (A + B) C = 180° - (24.49° + 44.77°) C = 180° - 69.26° C = 110.74° C A B 9.47 21.1 15.9 Florben G. Mendoza