Question

7. In a parallelogram PQRS, the bisectors of P and Q meet at O. Find POQ.​

Answer 1

`\mathrm{From\ figure,}\\\mathrm{\angle P+\angle Q=180^o}\ \ \ \ \mathrm{[Sum\ of\ cointerior\ angles\ is\ 180^o]}\\\mathrm{or,\ 2\angle OPQ+2 < OQP=180^o}\ \ \ \ \mathrm{[OP\ and\ OQ\ are\ angular\ bisectors]}\\\mathrm{or,\ \angle OPQ+\angle OQP=90^o.....(1)}\\\mathrm{In\ \triangle OPQ,}\\\mathrm{\angle POQ+\angle OPQ+\angle OQP=180^o\ \ \ \ [Sum\ of\ angles\ of\ triangle\ is\ 180^o.]}\\\mathrm{or,\ \angle POQ+90^o=180^o\ \ \ \ [From\ equation(1)]}\\\mathrm{\therefore \angle POQ=90^o}`