Question

In the given figure, O is the centre of a circle, BOA is its diameter and the tangent at the point P meets BA extended at T. If $\angle$ PBO = ${30}^o$ then $\angle$ PTA = ?

(a) ${60}^o$ (b) ${30}^o$ (c) ${15}^o$ (d) ${45}^o$

Answer 1

Join OP and radius is same
$\angle OPB=30$
$\angle POB=\mathrm{120......}$ [sum of triangle and inscribed angle property]
$\angle POT=\mathrm{60...}$[linear pair]
$\angle TPO=90$
then,
$\angle PTO+\angle POT+\angle TPO=180°$(sum of triangle property)
$\angle PTO=180°-90°-60°=30°$
$\angle PTO$$=\angle PTA=30°$is answer.

Correct option (b) 30°