Question

In figure $O$ is the center of the circle, $PQ$ is a chord and $PT$ is tangent to the circle at $P$. If $\angle POQ={70}^0$, find $\angle TPQ$.

Answer 1

$OP\cong OQ$(Radii of the same circle)

$\therefore \Delta POQ$ is an isosceles triangle

$\therefore \angle OPQ\cong \angle OQP$ ……………$\left(1\right)$

Let $\angle OPQ\cdot \angle OQP=x$ ………………$\left(2\right)$

In $\Delta OPQ$,

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

$x+x+{70}^o={180}^o$

$2x={180}^o-{70}^o$

$2x={110}^o$

$x={110}^o/2$

$\therefore x={55}^o$

PT is the tangent to the circle at P.

$\therefore OP\perp PT$

(tangent is perpendicular to radius)

$\therefore \angle OPT={90}^o$

$\angle OPT=\angle OPQ+\angle TPQ$

${90}^o={55}^o+\angle TPQ$

$\Rightarrow \angle TPQ={90}^o-{55}^o$

$\angle TPQ={35}^o$.