MCQ Questions on Geometry for Competitive Exams

$$\eqalign{ & PA = 12{\text{ cm}} \cr & \sin \theta = \frac{{AC}}{{AP}} = \frac{{AO}}{{OP}} \cr & \frac{{AC}}{{12}} = \frac{5}{{13}} \cr & AC = \frac{{60}}{{13}} \cr & \tan \theta = \frac{{AC}}{{PC}} = \frac{{AO}}{{AP}} \cr & \frac{{\frac{{60}}{{13}}}}{{PC}} = \frac{5}{{12}} \cr & PC = \frac{{144}}{{13}} \cr & {\text{Area of }}\Delta PCB = \frac{1}{2} \times PC \times AC \cr & {\text{Area of }}\Delta PAB = PC \times AC \cr & M = \frac{{144}}{{13}} \times \frac{{60}}{{13}} \cr & \frac{{\sqrt M }}{{15}} = \sqrt {\frac{{144 \times 60}}{{13 \times 13 \times 15}}} = \frac{{24}}{{13}} \cr} $$

$$x$$ = 180° - (60° + 28°) = 92°

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

AB = 6 cm, AC = 8 cm, BC = 10 cm AD ⊥ BC

AD = 5 cm