Question

if angle A and angle B are acute angle such that cosAis equal to cosB tben show that angle A =angle B

Answer 1

Let us suppose that ABQ is a right angle triangle such that triangle is right angled at B.

Given

          sinB = sinQ

Now

sinB = AQ/BQ ...........1

sinQ = AB/BQ ...........2

Since sinB = sinQ

then from equation 1 and 2, we get

      AQ/BQ = AB/BQ

=> AQ = AB

Therefore triangle ABQ is isoscales triangle.

 So Angle B = Angle Q

Hense proved.

Answer 2

Dear student
In △ABC, ∠C=90° and ∠A and ∠B are acute angle and by angle sum property of triangle,
∠A+∠B+∠C=180°
⇒∠A+∠B+90°=180°
⇒∠A+∠B=90°
⇒∠A=90°−∠B
Now,1)
sinA=sin(90°−B)=cosB⇒sinA=cosB2) cosA=cos(90°−B)=sinB⇒cosA=sinB