If in a triangle ABC, cosA cosB + sinA sinB sinC = 1. Then find the value of $\frac{a + b}{c}$ =

Open in App

Solution

$cos A . cos B + sin A . sin B . sin C = 1$
$∠ C = 90^{o}, ∠ A = ∠ B = 45^{o}$
$a = 2 R . sin A, b = 2 R . sin B, C = 2 R sin C$
$\Rightarrow \frac{a + b}{c} = \frac{2 R [\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}]}{2 R [1]}$
$= \frac{\sqrt{2}}{1} = \sqrt{2}$

Suggest Corrections

Similar questions

c o s A c o s B + s i n A s i n B s i n C = 1

a : b : c

△ A B C, s i n A + s i n B + s i n C = 1 + \sqrt{2}

c o s A + c o s B + c o s C = \sqrt{2}

sin A + sin B + sin C = 0

cos A + cos B + cos C = 0

cos (A + B) + cos (B + C) + cos (C + A) =

cos a + cos b + cos c = sin a + sin b + sin c = 0

cos (a - b) + cos (b - c) + cos (c - a) = - \frac{3}{2}

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)

Join BYJU'S Learning Program