Question

if A and B are acute angles such that sin a and Sin B then prove that angle A is equal to Angle B

Answer 1

Sin A = Sin B

=> BC / AB = AC / AB

=> BC = AC

We know that,

In a triangle the angles opposite to the equal sides are equal

So,

<A = <B

Hence Proved..

Answer 2

Answer:

Given:Sin A=Sin B

SinA=opp/hypo=BC/AB

SinB=opp/hypo=AC/AB

Two ratios are in a proportion in which the denominators are equal

hence in the given ratios AC=BC

In triangle ABC AC =BC

Therefore the triangle is an isosceles triangle

Hence angle A is equal to Angle B...

Hope this helps...