In Triangle ABC If $∠ B$ and in Triangle PQR $∠ Q$ are acute angles such that $sin B = sin Q$ then prove that $∠ B = ∠ Q$

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Solution

$A C = k . P R$
$A B = k . P Q$

$F r o m r i g h t △ A C B,$
$B y p y t h a g o r a s t h e o r e m w e h a v e,$

${A B}^{2} = {A C}^{2} + {B C}^{2}$
${(k . P Q)}^{2} = {(k . P R)}^{2} + {B C}^{2}$
$⟹ {B C}^{2} = k^{2} [{P Q}^{2} - {P R}^{2}]$

$∴ B C = k \sqrt{{P Q}^{2} - {P R}^{2}}$ ........ $(1)$

$F r o m r i g h t △ P R Q, b y p y t h a g o r a s t h m,$
${P Q}^{2} = {P R}^{2} + {Q R}^{2}$
$⟹ {Q R}^{2} = {P Q}^{2} - {P R}^{2}$

By putting in (1) we have $B C$ = $k Q R$

$h e n c e △ A C B \sim △ P R Q {S S S s i m i l a r i t y c r i t e r i a}$
$∵ ∠ B = ∠ Q$

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