Question

If $PQ$ is a tangent drawn from an external point $P$ to a circle with center $O$ and $QOR$ is a diameter where the length of $QOR$ is $8$ cm such that $\angle POR={120}^0,$ then find OP and PQ.

Answer 1

We are given that,

Circle with centre O is having diameter $QR=8$ cm.

Now, $\angle POQ$ and $\angle POR$ is making linear pair.

So, $m\angle POQ+m\angle POR={180}^{\circ }$

$m\angle POQ={180}^{\circ }-{120}^{\circ }$

$={60}^{\circ }$

Now, $OR=OQ=$ radius $=\frac{8}{2}=4$ cm

& $\Delta OPQ$ is right angle triangle.

So, $\sin {60}^{\circ }=\frac{PQ}{OP}$

$\frac{\sqrt{3}}{2}=\frac{PQ}{OP}⟹\frac{PQ}{\sqrt{3}}=\frac{OP}{2}=k$

Now,

$O{P}^2=Q{O}^2+P{Q}^2$

$(2k{)}^2=(4{)}^2+(\sqrt{3}k{)}^2$

$4k^2=16+3k^2$

$k^2=16⟹k=\pm 4$

Now, sides are positive. So, $PQ=4\sqrt{3}$ cm & $OP=8$ cm