Question

Mark (✓) against the correct answer in each of the following:
In the given figure, O is the centre of a circle and ∠BAC = 40°. Then, ∠OBC = ?
(a) 40°
(b) 50°
(c) 80°
(d) 20°

Answer 1

(b) 50°

$$\begin{gathered} \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{angle}\mathrm{subtended}\mathrm{by}\mathrm{an}\mathrm{arc}\mathrm{of}\mathrm{a}\mathrm{circle}\mathrm{at}\mathrm{the}\mathrm{centre} \\ \mathrm{is}\mathrm{double}\mathrm{the}\mathrm{angle}\mathrm{subtended}\mathrm{by}\mathrm{it}\mathrm{at}\mathrm{any}\mathrm{point}\mathrm{on}\mathrm{the}\mathrm{circumference}\mathrm{of}\mathrm{the}\mathrm{circle}. \\ \mathrm{So},\angle BOC=2\angle BAC \\ \Rightarrow \angle BOC=2\times 40°=80° \\ \mathrm{Now},BO=OC. \\ \mathrm{So},∆BOC\mathrm{is}\mathrm{an}\mathrm{isoceles}\mathrm{triangle}. \\ \mathrm{In}∆BOC, \\ \angle BOC+\angle OBC+\angle BCO=180°(\text{A}\mathrm{ngle}-\mathrm{sum}\mathrm{property}) \\ \Rightarrow 80°+\angle OBC+\angle OBC=180°(\angle OBC=\angle BCO) \\ \Rightarrow 2\angle OBC=180°-80° \\ \Rightarrow 2\angle OBC=100° \\ \Rightarrow \angle OBC=50° \end{gathered}$$