Question

In Triangle ABC If $$\angle{B}$$ and in Triangle PQR   $$\angle{Q}$$  are acute angles such that $$\sin B=\sin Q$$ then prove that $$\angle B=\angle Q$$

Solution

$$AC=k.PR$$$$AB=k.PQ$$$$From\quad right\triangle ACB,$$$$By\quad pythagoras\quad theorem\quad we\quad have,$$$${ AB }^{ 2 }={ AC }^{ 2 }+{ BC }^{ 2 }$$$${ (k.PQ) }^{ 2 }={ (k.PR) }^{ 2 }+{ BC }^{ 2 }$$$$\implies\quad { BC }^{ 2 }={ k }^{ 2 }[{ PQ }^{ 2 }-{ PR }^{ 2 }]$$$$\therefore BC=k\sqrt { { PQ }^{ 2 }-{ PR }^{ 2 } }$$........$$(1)$$$$From\quad right \ \triangle PRQ,by\quad pythagoras\quad thm,$$$${ PQ }^{ 2 }={ PR }^{ 2 }+{ QR }^{ 2 }$$$$\implies\quad { QR }^{ 2 }={ PQ }^{ 2 }-{ PR }^{ 2 }$$ By putting in (1) we have $$BC\,$$=$$\,k\,QR$$$$hence\triangle ACB\sim \triangle PRQ\quad \{ SSS\quad similarity\quad criteria\}$$$$\because \angle B=\angle Q$$Mathematics

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