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