Question

If $\angle B$ and $\angle Q$ are acute angels such that $\sin B-\sin Q$, then prove that $\angle B=\angle Q$.

Answer 1

We have,

In $\Delta ABCand\Delta PQR$

Given that, $\angle Band\angle Q$ are acute angles

$\therefore AC=k\times PRandAB=k\times PQ$

From $\Delta ACB$

By Pythagoras theorem,

$A{B}^2=A{C}^2+B{C}^2$

$\Rightarrow {(k\times PR)}^2={(k\times PQ)}^2+B{C}^2$

$\Rightarrow k^2\times P{R}^2=k^2\times P{Q}^2+B{C}^2$

$\Rightarrow B{C}^2=k^2\times P{R}^2-k^2\times P{Q}^2$

$\Rightarrow B{C}^2=k^2[P{R}^2-P{Q}^2]$

$\Rightarrow BC=k\sqrt{P{R}^2-P{Q}^2}$

From right angle triangle PRQ,

Using Pythagoras theorem,

We have,

$P{Q}^2=P{R}^2+Q{R}^2$

$\Rightarrow Q{R}^2=P{Q}^2-P{R}^2$

Hence,

$\Delta ACB\sim \Delta PRQ(S.S.S.similaritytriangle)$

$\therefore \angle B=\angle Q$

Hence, this is the answer.