Question

If $O$ is the centre of a circle, $PQ$ is a chord and the tangent $PR$ at $P$ makes an angle of ${50}^0$ with $PQ$ , then find the Angle$\left(POQ\right)$.

Answer 1

Given that, $O$ is the centre of a circle, $PQ$ is a chord and the tangent $PR$ at $P$ makes an angle of ${50}^o$ with $PQ$.

We need to find $\angle POQ$.

We know that the tangent is perpendicular to the radius.

$\therefore \angle OPQ+\angle QPR={90}^o$

From the figure $\angle QPR={50}^o$.

$\Rightarrow \angle OPQ+{50}^o={90}^o$

$\Rightarrow \angle OPQ={90}^o-{50}^o$

$\therefore \angle OPQ={40}^o$

We know that, the angles opposite to the equal sides of the triangle are equal.

$\therefore \angle OPQ=\angle OQP={40}^o$

Also, we know that sum of angles in the triangle is ${180}^o$.

$\Rightarrow \angle OPQ+\angle OQP+\angle POQ={180}^o$

$\Rightarrow {40}^o+{40}^o+\angle POQ={180}^o$

$\Rightarrow {80}^o+\angle POQ={180}^o$

$\Rightarrow \angle POQ={180}^o-{80}^o$

$\therefore \angle POQ={100}^o$