Question

in triangleabcbisector of angle A and angle B intersect at point O if angel C is 70 find measure of angle AOB

Answer 1

Sin^2 A = 1/2 tan^2A= 1/(2*(1)^2) =1/2, As angle A is an acute angle.

So,

SinA = 1/√2 , You know value of sin 45°=1/√2

Or,

A = Sin^(-1) {1/√2} = 45°

So, A = 45°