Question

$BOA$ is the diameter of a circle and the tangent at point $P$ meets $AB$ produced at $T$. If $\angle PBO={30}^o$, find $\angle PTA$.

Answer 1

Given, $PT$ is a tangent

$\angle PBA={30}^0$

Now,

$\angle BPA={90}^0$ (angle in semicircle)

$\angle OPT={90}^0$ (angle between the radius and tangent)

Also in $\Delta BPA$

$\angle PAB={180}^0-(\angle {30}^0+\angle BPA)$

$={180}^0-{30}^0-{90}^0$

$={60}^0⟶\left(1\right)$

Now look into the $\Delta OPA$,

$OP=OA$ (radius of circle)

$\Rightarrow \angle OAP=\angle OPA={60}^0$ $(\because \angle OAP=\angle BAP={60}^0)$

$\therefore$ $\angle APT=\angle OPT-\angle OPA={90}^0-{60}^0={30}^0$

$\therefore$ $\angle PAT={180}^0-\left({60}^0\right)={120}^0$ (supplementary angle)

In $\Delta PAT$,

$\angle PAT+\angle PTA+\angle APT={180}^0$

$\Rightarrow {120}^0+\angle PTA+{30}^0={180}^0$

$\Rightarrow \angle PTA={30}^0$.