Question

If ∠ B and ∠ Q are acute angles such that sin B = sin Q, then prove that ∠ B = ∠ Q.

Answer 1

Let us consider two right triangles ABC and PQR where sin B = sin Q 
we have                sin B = AC/AB 
and                        sin Q = PR/PQ
                            AC/AB = PR/PQ
then,                   AC/PR = AB/PQ = k , say _____________  (i)
Now Using Pythagoras Theorem ,
                            BC = √AB² - AC²
                             QR = √PQ² - PR²
So, BC/QR = √AB² - AC² / √PQ² - PR² = √K²PQ² - K²PR² / √PQ² - PR² = k _____  (ii)
FROM  (i) and (ii) , we have
                            AC/PR = AB/PQ = BC/QR
Then by using theorem ΔACB ≈  ΔPRQ And therefore <B = < Q