O is a point in the interior of ΔABC such that OA = 12 cm, OC = 9 cm, ∠AOB = ∠BOC = ∠COA and ∠ABC = 60°. What is the length (in cm) of OB?

Solution (By Examveda Team)

$$\eqalign{ & {\text{In }}\Delta AOB \cr & \frac{{\sin \theta }}{{OB}} = \frac{{\sin \left( {{{60}^ \circ } - \theta } \right)}}{{AO}} \cr & \frac{{\sin \theta }}{{\sin \left( {{{60}^ \circ } - \theta } \right)}} = \frac{{OB}}{{12}}{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {\text{i}} \right) \cr & {\text{In }}\Delta BOC \cr & \frac{{\sin \theta }}{{OC}} = \frac{{\sin \left( {{{60}^ \circ } - \theta } \right)}}{{OB}} \cr & \frac{{\sin \theta }}{{\sin \left( {{{60}^ \circ } - \theta } \right)}} = \frac{9}{{OB}}{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{ii}}} \right) \cr & {\text{Compare }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & \frac{{OB}}{{12}} = \frac{9}{{OB}} \cr & O{B^2} = 12 \times 9 \cr & O{B^2} = 108 \cr & OB = 6\sqrt 3 \cr} $$