Question

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

Answer 1

$$\begin{gathered} \text{Given,}BC\text{and}BD\text{are the two tangents drawn to the circle with centre O, such that}\angle CBD={120}^0. \\ \text{They are equally inclined to the line segment joining the centre to that point.} \\ \therefore \angle OBC=\angle OBD=\frac{1}{2}\angle CBD={60}^0 \\ \text{Now, tangent drawn from an external point is perpendicular to the radius} \\ \text{at the point of contact}\text{.} \\ \therefore \angle OCB=\angle ODB={90}^0 \\ \begin{array}{l}\therefore \text{From}\Delta OBC,\angle BOC={180}^0-(\angle OCB+\angle OBC)\\ \Rightarrow \angle BOC={180}^0-({90}^0+{60}^0)\\ \Rightarrow \angle BOC={30}^0.\end{array} \\ \begin{array}{l}\text{Now from right-angled}△OBC,\frac{BC}{OB}=\sin {30}^0\\ \Rightarrow \frac{BC}{OB}=\frac{1}{2}\\ \Rightarrow OB=2BC.\end{array} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$