Question

In $△PQR$, $\angle P={30}^o$, $\angle Q={60}^0$, $\angle R={90}^o$ and $PQ=10$ units.

Find $PR$ and $QR$.

• A. $PR=$ $10$ units, $QR=$ $5\sqrt{3}$ units
• B. $PR=$ $5$ units, $QR=$ $5\sqrt{3}$ units
• C. $PR=$ $5\sqrt{3}$ units, $QR=$ $5$ units
• D. $PR=$ $5$ units, $QR=$ $10\sqrt{3}$ units

Answer 1

Answer: (C)

$PR=$ $5\sqrt{3}$ units, $QR=$ $5$ units

$△PQR$ is a ${30}^o-{60}^o-{90}^o$ triangle ....given

Since, $\angle R={90}^o$, side $PQ$ is hypotenuse.

$\Rightarrow$ By ${30}^o-{60}^o-{90}^o$ theorem,

$\Rightarrow PR=\frac{\sqrt{3}}{2}\times PQ$ and $QR=\frac{1}{2}\times PQ$

$\Rightarrow PR=\frac{\sqrt{3}}{2}\times 10$ and $QR=\frac{1}{2}\times 10$

$\Rightarrow PR=5\sqrt{3}$ units and $QR=5$ units

So, option C is correct.