Question

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Here is question for you . iit level + Questions
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If in a triangle ABC cosAcosB + sinA sinB sinC = 1 then proove that a : b : c =1 : 1 : √2
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Answer 1

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We are given that in ∆ ABC ;

cos A cos B + sin A sin B sin C = 1

⇒ sin A sin B sin C = 1 – cos A cos B

⇒ sin C = 1 – cos A cos B/sin A sin B

⇒ 1 – cos A cos B/sin A sin B ≤ 1 [∵ sin C ≤ 1]

⇒ 1 – cos A cos B ≤ sin A sin B

⇒ 1 ≤ cos A cos B + sin A sin B

⇒ 1 ≤ cos(A – B)

⇒ 1 ≤ cos(A – B)

But we know cos (A – B) ≤1

∴ We must have cos (A – B) = 1

⇒ A – B = 0

⇒ A = B

∴ cos A cos A + sin A sin A sin C = 1 [For A = B]

⇒ cos2 A + sin2 A sin C = 1

⇒ sin2 A sin C = 1 – cos2 A

⇒ sin2 A sin C = sin2 A

⇒ sin2 A (sin C – 1) = 0

⇒ sin A = 0 or sin C = 1

The only possibility is sin C = 1 ⇒ C = π/2

∴ A + B = π/2

But A = B ⇒ A = B = π/4

∴ By Sine law in ∆ ABC,

a/sin A = b/sin B = c/sin C

⇒ a/sin 45° = b/sin 45° = c/sin 90°

⇒ a/1/√2 = b/1/√2 = c/1

⇒ a/1 = b/1 = 1/√2 ⇒ a : b : c = 1 : 1 : √2







Hence proved the result

Answer 2

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