Question

Two tangents BC and BD are drawn to a circle with centre O, such that ∠CBD = 120°. Prove that OB = 2BC.

Answer 1

$$\begin{gathered} \mathrm{Here},OB\mathrm{is}\mathrm{}\mathrm{the}\mathrm{}\mathrm{bisector}\mathrm{}\mathrm{of}\angle CBD. \\ (\text{T}\mathrm{wo}\mathrm{tangents}\mathrm{are}\mathrm{equally}\mathrm{inclined}\mathrm{to}\mathrm{the}\mathrm{line}\mathrm{segment}\mathrm{joining}\mathrm{the}\mathrm{centre}\mathrm{to}\mathrm{that}\mathrm{point}) \\ \therefore \angle CBO=\angle DBO=\frac{1}{2}\angle CBD={60}^0 \\ \therefore \mathrm{From}\mathrm{}∆BOD,\angle BOD={30}^0 \\ \mathrm{Now},\mathrm{from}\mathrm{}\mathrm{right}-\mathrm{angled}∆BOD, \\ \frac{BD}{OB}=\sin {30}^0 \\ \Rightarrow \frac{BD}{OB}=\frac{1}{2} \\ \Rightarrow OB=2BD \\ \Rightarrow OB=2BC\left(\text{S}\mathrm{ince}\mathrm{}\mathrm{tangents}\mathrm{}\mathrm{from}\mathrm{}\mathrm{an}\mathrm{}\mathrm{external}\mathrm{}\mathrm{point}\mathrm{}\mathrm{are}\mathrm{}\mathrm{equal},\mathrm{i}.\mathrm{e}.BC=BD\right) \\ \therefore OB=2BC \end{gathered}$$