Question

In the figure, PQ is a chord of a circle with center O and PT is a tangent. If $\angle QRT={60}^o$, find $\angle PSQ$.

Answer 1

Let $y$ be the on the tangent $PT$.

$\angle QPT+\angle QPX={180}^o\left(\text{From}\text{linear}\text{pair}\right)$

$\Rightarrow \angle QPX={180}^o-\angle QPT$

$\Rightarrow \angle QPX={180}^o-{60}^o$

$\Rightarrow \angle QPX={120}^o$

So,

$\angle QPX=\angle PRQ$

$\Rightarrow \angle PRQ={120}^o$

We know that,

$\angle POQ+\angle PRQ={180}^o$

$\Rightarrow \angle POQ={180}^o-{120}^o$

$\Rightarrow \angle POQ={60}^o$

Using circle triangle theorem,

$\angle POQ=2\angle PSQ$

${60}^o=2\angle PSQ$

$\angle PSQ={30}^o$

Hence, this is the answer.