Question

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

Answer 1

Hint: For the above question we will first assume two different right angled triangle say ABC and PQR and then we will use the concept that sine of any angle is equal to the ratio of perpendicular to hypotenuse of that right angled triangle and we will get the required result.

Complete step-by-step answer:
We have been given that $\angle B\ and\ \angle Q$ are acute angles such that $\sin B=\sin Q$ and we need to prove $\angle B=\angle Q$.
Let us suppose $\Delta ABC\ and\ \Delta PQR$ right angled at C and R respectively.

In $\Delta ABC$,
$\sin B=\dfrac{AC}{AB}$
In $\Delta PQR$,
$\sin Q=\dfrac{PR}{PQ}$
Since, sine of any angle is the ratio of perpendicular to hypotenuse.
Now, according to question we have,
$\begin{aligned} & \sin B=\sin Q \\\ & \Rightarrow \dfrac{AC}{AB}=\dfrac{PR}{PQ} \\\ & \Rightarrow \dfrac{AC}{PR}=\dfrac{AB}{PQ} \\\ & Let,\dfrac{AC}{PR}=\dfrac{AB}{PQ}=K.........\left( 1 \right) \\\ \end{aligned}$
Now, in $\Delta ABC$, using Pythagoras theorem we have,
$BC=\sqrt{A{{B}^{2}}-A{{C}^{2}}}$
Using (i) we get $AB=K.PQ\ and\ AC=K.PR$.

& \Rightarrow BC=\sqrt{{{\left( K.PQ \right)}^{2}}-{{\left( K.PR \right)}^{2}}} \\\ & =\sqrt{{{K}^{2}}P{{Q}^{2}}-{{K}^{2}}P{{R}^{2}}} \\\ & =\sqrt{{{K}^{2}}\left( P{{Q}^{2}}-P{{R}^{2}} \right)} \\\ & =K\sqrt{P{{Q}^{2}}-P{{R}^{2}}}...........\left( 2 \right) \\\ \end{aligned}$$ In $\Delta PQR$, using Pythagoras Theorem we have, $QR=\sqrt{P{{Q}^{2}}-P{{R}^{2}}}$ From equation (2) we have $BC=K\sqrt{P{{Q}^{2}}-P{{R}^{2}}}$. $\Rightarrow \dfrac{BC}{QR}=\dfrac{K\sqrt{P{{Q}^{2}}-P{{R}^{2}}}}{\sqrt{P{{Q}^{2}}-P{{R}^{2}}}}=K............\left( 3 \right)$ On comparing (1) and (3), we get, $\dfrac{AC}{PR}=\dfrac{AB}{PQ}=\dfrac{BC}{QR}$ So, by using SSS a similar condition which states that two triangles are similar if their corresponding sides are proportional. $\Rightarrow \Delta ABC\sim \Delta PQR$ According to corresponding part of similar triangle we have, $\angle B=\angle Q$ Hence, it is proved. Note: Remember that Pythagoras Theorem states that in a right angled triangle the sum of square of perpendicular and base is equal to square of hypotenuse of the triangle. While applying it in the solution, we must not make any mistake in the ratio of sides. Also, remember the property of a similar triangle that two similar triangles have corresponding angles are congruent and their corresponding sides are in proportion.