If in a triangle ABC,if sinA : sinB : sinC =4 : 5 : 6, then prove that, cosA,: cosB : cosC=12 : 9 : 2.(
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Given If in a triangle ABC,if sinA : sinB : sinC = 4 : 5 : 6, then prove that, cosA,: cosB : cosC = 12 : 9 : 2
- So given sin A : sin B : sin C = 4:5:6
- Let sin A = 4p, sin B = 5p, sin C = 6p
- So a / sin A = b/sin B = c/sin C = k (where k is a constant)
- So a = k.sin A
- Or a = k.4xp
- Similarly b = k sin B
- Or b = k.5p
- So c = k. sin C
- Or c = k. 6p
- Now let k.p = m
- So a = 4m, b = 5m, c = 6m
- Now cos A = b^2 + c^2 – a^2 / 2bc
- = 25m^2 + 36 m^2 – 16 m^2 / 2 (5m) (6m)
- = m^2 (45) / 60 m^2
- Cos A = ¾
- Cos B = a^2 + c^2 – b^2 / 2ac
- = 16m^2 + 36 m^2 – 25 m^2 / 2 (4m)(6m)
- = 27 m^2 / 48 m^2
- = 27 / 48
- Cos B = 9 / 16
- Cos C = a^2 + b^2 – c^2 / 2ab
- = 16 m^2 + 25 m^2 – 36 m^2 / 2 (4 m) (5m)
- = 5 m^2 / 40 m^2
- Cos C = 1/8
- So we get
- Cos A = ¾, Cos B = 9/16 and Cos C = 1/8
- Multiplying by 16 we get
- 16 Cos A
- 16 x ¾
- 12
- 16 Cos B
- 16 x 9/16
- 9
- 16 Cos C
- 16 x 1/8
- 2
- So 16 Cos A : 16 Cos B : 16 Cos C = 12 : 9 : 2
- Or Cos A : Cos B : Cos C = 12 : 9 : 2 (proved)
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https://brainly.in/question/18227318
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Answer:
Upper one is perfect ........
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