Question

In $△PQR,\frac{PQ}{1}=\frac{PR}{2}=\frac{QR}{\sqrt{3}}$, then $m\angle R=$

• A. ${90}^o$
• B. ${60}^o$
• C. ${45}^o$
• D. ${30}^o$

Answer 1

The correct option is D ${30}^o$

In $△PQR,\frac{PQ}{1}=\frac{PR}{2}=\frac{QR}{\sqrt{3}}$

Here, $PQ:PR:QR=1:2:\sqrt{3}$

Let $PQ=x⟹PR=2x$ and $QR=\sqrt{3}x$

Now, ${PQ}^2+{QR}^2=x^2+3x^2=4x^2=(2x{)}^2={PR}^2$

$\therefore$ $PR$ is the hypotenuse

$⟹m\angle Q={90}^o$

Let $\angle R=\theta$

$⟹\sin \theta =\frac{PQ}{PR}=\frac{x}{2x}=\frac{1}{2}$

Then $\sin \theta =\frac{1}{2}⟹\theta ={\sin }^{-1}\left(\frac{1}{2}\right)$
$\therefore \theta ={30}^o$