Module 8 Lesson 5 Oblique Triangles Florben G. Mendoza

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

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

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

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

S S A CASE 2: SSA Florben G. Mendoza

S S A CASE 3: SAS Florben G. Mendoza

S S S CASE 4: SSS Florben G. Mendoza

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

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

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

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

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

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 = SAA Florben G. Mendoza

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 = Florben G. Mendoza

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 = ASA Florben G. Mendoza

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 = Florben G. Mendoza

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

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

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    Ambiguous Case (SSA) Florben G. Mendoza

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

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

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 B = 45.23° SSA Florben G. Mendoza

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° °) C = 180° ° C = 94.77° a sin A = c sin C 9 sin 40° = c sin 94.77° c (sin 40°) = 9 (sin 94.77°) c = Step 4: (sin 40°) Florben G. Mendoza

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

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

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 = b < h ( No Solution) A C B b a c h Florben G. Mendoza

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 A = 48.59° Step 2: B = 180° - (A + C) B = 180° - (48.59° + 100°) B = 180° ° B = 31.41° C A B 100° b Florben G. Mendoza

29 Example 5: SSA Let C = 100°, a = 25, and c = 33 C A B 100° 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 = Florben G. Mendoza

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

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 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 Law of Cosines Similarly Note the pattern Florben G. Mendoza

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 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 = – a 2 = a =10.41 A B C 42° a Florben G. Mendoza

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° a cos B = cos B = 0.56 B = cos B = 55.94° SAS Florben G. Mendoza

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

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 = cos A = 0.91 A = cos A = 24.49° Florben G. Mendoza

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 = cos B = 0.71 B = cos B = 44.77° Florben G. Mendoza

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