Question

If PA and PB are two tangents to a circle with centre O, such that ∠AOB = 110°, find ∠APB.


(a) 90°
(b) 60°
(c) 70°
(d) 55°

Answer 1

(c) 70°
$$\begin{gathered} \mathrm{Given},PA\mathrm{and}PB\mathrm{are}\mathrm{}\mathrm{tangents}\mathrm{}\mathrm{to}\mathrm{}\mathrm{a}\mathrm{}\mathrm{circle}\mathrm{}\mathrm{with}\mathrm{}\mathrm{centre}O,\mathrm{with}\angle AOB={110}^0. \\ \mathrm{Now},\mathrm{}\mathrm{we}\mathrm{}\mathrm{know}\text{that}\mathrm{tangents}\mathrm{}\mathrm{drawn}\mathrm{}\mathrm{from}\mathrm{}\mathrm{an}\mathrm{}\mathrm{external}\mathrm{}\mathrm{point}\mathrm{}\mathrm{are}\mathrm{}\mathrm{perpendicular} \\ \mathrm{to}\mathrm{}\mathrm{the}\mathrm{}\mathrm{radius}\mathrm{}\mathrm{at}\mathrm{}\mathrm{the}\mathrm{}\mathrm{point}\mathrm{}\mathrm{of}\mathrm{}\mathrm{contact}. \\ \mathrm{So},\angle OAP={90}^{0}and\angle OBP={90}^0. \\ \Rightarrow \angle OAP+\angle OBP={90}^0+{90}^0={180}^0,\mathrm{which}\mathrm{}\mathrm{shows}\mathrm{}\mathrm{that}OABP\mathrm{}\mathrm{is}\mathrm{}\mathrm{a}\mathrm{}\mathrm{cyclic} \\ \mathrm{quadrilateral}. \\ \therefore \angle AOB+\angle APB={180}^0 \\ \Rightarrow {110}^0+\angle APB={180}^0 \\ \Rightarrow \angle APB={180}^0-{110}^0 \\ \Rightarrow \angle APB={70}^0 \end{gathered}$$