If in a $△ A B C,$ $(sin A + sin B + sin C) (sin A + sin B - sin C) = 3 sin A sin B,$ then angle $C$ (in degree) is

Open in App

Solution

$(sin A + sin B + sin C) (sin A + sin B - sin C) = 3 sin A sin B \Rightarrow (sin A + sin B)^{2} - {sin}^{2} C = 3 sin A sin B \Rightarrow {sin}^{2} A + {sin}^{2} B - {sin}^{2} C = sin A sin B \Rightarrow {sin}^{2} A + sin (B + C) sin (B - C) = sin A sin B [∵ A + B + C = π] \Rightarrow {sin}^{2} A + sin (π - A) sin (B - C) = sin A sin B \Rightarrow sin A [sin (A) + sin (B - C)] = sin A sin B \Rightarrow sin A [sin (B + C) + sin (B - C)] = sin A sin B \Rightarrow sin A (2 sin B cos C) = sin A sin B \Rightarrow sin A sin B (2 cos C - 1) = 0 \Rightarrow cos C = \frac{1}{2} \Rightarrow C = 60^{\circ} [∵ sin A sin B \neq 0]$

Suggest Corrections

Similar questions

A B C

(sin A + sin B + sin C) (sin A + sin B - sin C) = 3 sin A sin B

A + B + C = π

(sin A + sin B + sin C) (- sin A + sin B + sin C) (sin A - sin B + sin C) (sin A + sin B - sin C) = 4 {sin}^{2} A {sin}^{2} B {sin}^{2} C

A, B, C

△ A B C

(s i n A + s i n B) (s i n B + s i n C) (s i n C + s i n A)

Δ

A B C

A, B, C

a, b, c

(sin A + sin B + sin C) (sin A + sin C - sin B) = μ sin A sin C

sin A = a k, sin B = b k, sin C = c k

μ

\sum \frac{sin (C - A)}{sin C sin A}

sin A, sin B, sin C

Join BYJU'S Learning Program