Question

O is the centre of the circle. Prove that Q₁ + P = 90°.

Answer 1

`\mathrm{Solution:}\\\mathrm{Given:\ O\ is\ the\ centre\ of\ circle.}\\\mathrm{To\ prove:\ \angle OQR+\angle P=90^o}\\\mathrm{Proof:}\\`

`\mathrm{(i).\\ \ \angle OQR=\angle ORQ\ \ \ \ [\mathrm{OQ=OR,\ radii\ of\ same\ circle.}]}\\\mathrm{(ii)\ \angle OQR+\angle ORQ+\angle QOR=180^o\ \ \ [\mathrm{Sum\ of\ angles\ of\ triangle\ is\ 180^o.}]}\\\mathrm{(iii)\ \angle QOR=2\angle P\ \ \ [Circular\ angle\ is\ twice\ the\ inscribed\ angle\ on\ same\ arc.]}\\`

`\mathrm{(iv)\ \angle OQR +\angle OQR+\angle QOR=180^o\ \ [From\ statement\ 1,\ \angle OQR=\angle ORQ}]\\\mathrm{or,\ }\mathrm{2 \angle OQR+2\angle P=180^o\ \ \ [From\ statement\ (iii).]\\}\\\mathrm{\therefore \angle OQR+\angle P=90^o}`