CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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 $$\angle B=\angle Q$$ 
1002060_01b90fd12deb4dca9251e9168b9e5334.PNG


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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More



footer-image