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# If in a triangle ABC, cos A cos B + sin A sin B sin C = 1, Show that a : b : c = 1 : 1 : √2

Jitender Pal
7 years ago
Hello Student,
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.

Thanks
Jitender Pal