In the given figure, PQRS is a square of side 8 cm. ∠PQO = 60°. What is the area (in cm²) of the triangle POQ?

Solution (By Examveda Team)

$$\eqalign{ & {\text{In }}\sin \,{\text{rule}} \cr & \frac{{\sin {{75}^ \circ }}}{{PQ}} = \frac{{\sin {{60}^ \circ }}}{{PO}} \cr & \frac{{\sin \left( {{{30}^ \circ } + {{45}^ \circ }} \right)}}{8} = \frac{{\sin {{60}^ \circ }}}{{PO}} \cr & \frac{{\frac{1}{2} \times \frac{1}{{\sqrt 2 }} + \frac{{\sqrt 3 }}{2} \times \frac{1}{{\sqrt 2 }}}}{8} = \frac{{\sqrt 3 }}{{2PO}} \cr & \frac{{\sqrt 3 + 1}}{{2\sqrt 2 \times 8}} = \frac{{\sqrt 3 }}{{2PO}} \cr & PO = \frac{{8\sqrt 6 }}{{\left( {\sqrt 3 + 1} \right)}} \cr & {\text{After rationaltion}} \cr & PO = \frac{{8\sqrt 6 }}{{\left( {\sqrt 3 + 1} \right)}} = 4\sqrt 6 \left( {\sqrt 3 - 1} \right) \cr & {\text{Then in }}\Delta POD, \cr & {\text{Area}} = \frac{1}{2}PQ \times PO\sin {45^ \circ } \cr & = \frac{1}{2} \times 8 \times 4\sqrt 6 \left( {\sqrt 3 - 1} \right) \times \frac{1}{{\sqrt 2 }} \cr & = 2\sqrt 2 \times 4\sqrt 6 \left( {\sqrt 3 - 1} \right) \cr & = 16\left( {3 - \sqrt 3 } \right) \cr} $$