Question

If $\angle A$ and $\angle P$ are acute angles such that $\sin A=\sin P$ then prove that $\angle A=\angle P$

Answer 1

Given $\sin A=\sin P$

$\therefore \frac{BC}{AC}=\frac{RQ}{PR}$

Let $\frac{BC}{RQ}=\frac{AC}{PR}=k$

$BC=kRQ$ and $AC=kPR$

Now,

$\frac{AB}{PQ}=\frac{\sqrt{A{C}^2-B{C}^2}}{\sqrt{P{R}^2-R{Q}^2}}$

$\frac{AB}{PQ}=\frac{\sqrt{k^{2}P{R}^2-k^{2}R{Q}^2}}{\sqrt{P{R}^2-R{Q}^2}}$

$\frac{AB}{PQ}=\frac{k\sqrt{P{R}^2-R{Q}^2}}{\sqrt{P{R}^2-R{Q}^2}}$

$\frac{AB}{PQ}=k$

$\therefore \frac{AB}{PQ}=\frac{BC}{RQ}=\frac{AC}{PR}$

$\therefore △ABC\sim △PQR$ [By SSS similarity]

$\therefore \angle A=\angle P$