Question

In $△PQR$, $\angle P={30}^0,\angle Q={60}^0,\angle R={90}^0$ and $PQ=10$, then find $PR$ and $QR$.

Answer 1

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

$\sin 60=\frac{PR}{PQ}$ ( $\sin \theta =\frac{Perpendicular}{Hypotenuse}$)

$\frac{\sqrt{3}}{2}=\frac{PR}{10}$

$PR=5\sqrt{3}$ cm

$\cos 60=\frac{QR}{PQ}$ ( $\cos \theta =\frac{Base}{Hypotenuse}$)

$\frac{1}{2}=\frac{QR}{10}$

$QR=5$ cm