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

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by rajender243

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by rajender243

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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