Question

In figure, PQ is a tangent from an external point P to a circle with centre O and OP cuts the circle at T and QOR is a diameter. If $\angle POR={130}^{\circ }$ and S is a point on the circle, find $\angle 1+\angle 2$

Answer 1

Given: $\angle POR={130}^{\circ }$

$\therefore \angle POR=\angle ROT=2\angle RST$ (angle at center = 2 times angle at cicumference)
$\Rightarrow {130}^{\circ }=2\angle RST$
$\Rightarrow \angle RST=\angle 2={65}^{\circ }$

$\angle OQP={90}^{\circ }$ (The point of contact of tangent and radius)
$\angle 1+{90}^{\circ }={130}^{\circ }$ (since exterior angle=sum of two opposite interior angles in a triangle)
$\angle 1={40}^{\circ }$

$\angle 1+\angle 2={65}^{\circ }$ + ${40}^{\circ }$ = ${105}^{\circ }$

Hence, $\angle 1+\angle 2={105}^{\circ }$