Question

In Fig. 8.79, 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 ∠POR = 130° and S is a point on the circle, find ∠1 + ∠2.

Answer 1

Given: ∠POR = 130°
So, ∠TSR = \[\frac{1}{2}\angle POR = \frac{1}{2} \times 130^o = 65^o = \angle2\] (Since angle subtended by the arc at the centre is double   the angle subtended by it at the remaining part of the circle)
∠POQ = 180º − ∠POR = 180º − 130º = 50º      .....(2)          (Linear pair)       
In

\[\frac{1}{2}\angle POR = \frac{1}{2} \times 130^o = 65^o = \angle2\]

Δ POQ, \[\angle1 + \angle POQ + \angle OQP = 180^o\]
\[ \Rightarrow \angle1 + 50^o + 90^o = 180^o\]
\[ \Rightarrow \angle1 = 40^o\]

\[\angle1 + \angle POQ + \angle OQP = 180^o\]
\[ \Rightarrow \angle1 + 50^o + 90^o = 180^o\]
\[ \Rightarrow \angle1 = 40^o\]

\[Now \angle1 + \angle2 = 40^o + 65^o = 105^o\]