in the adjoining figure PQ and PR are tangents from P to the circle with Centre O if PO R equals to 55 degree find QPR

Dear Student,

Consider the figure given below:


Since, PQ and PR are the tangents to the circle with centre 'O'.

So, $\mathrm{PQ}\perp \mathrm{OQ} \mathrm{and} \mathrm{PR}\perp \mathrm{OR}$         [The tangent to a circle is perpendicular to the radius through the point of contact.]
$$\begin{gathered} \mathrm{In} ∆\mathrm{ORP}, \\ \angle \mathrm{POR} + \angle \mathrm{ORP} + \angle \mathrm{OPR} = 180° \left[\mathrm{Angle} \mathrm{Sum} \mathrm{Property}\right] \\ \Rightarrow 55° + 90° + \angle \mathrm{OPR} = 180° \\ \Rightarrow 145° + \angle \mathrm{OPR} = 180° \\ \Rightarrow \angle \mathrm{OPR} = 180° - 145° = 35° \end{gathered}$$

$$\begin{gathered} \mathrm{In} ∆\mathrm{OQP} \mathrm{and} ∆\mathrm{ORP} \\ \mathrm{OQ} = \mathrm{OR} [\mathrm{radii} \mathrm{of} \mathrm{circle}] \\ \angle \mathrm{OQR} = \angle \mathrm{ORP} = 90° \left[\mathrm{PQ}\perp \mathrm{OQ} \mathrm{and} \mathrm{PR}\perp \mathrm{OR}\right] \\ \mathrm{PQ} = \mathrm{PR} \left[\mathrm{The} \mathrm{lengths} \mathrm{of} \mathrm{the} \mathrm{two} \mathrm{tangents} \mathrm{from} \mathrm{an} \mathrm{external} \mathrm{point} \mathrm{to} \mathrm{a} \mathrm{circle} \mathrm{are} \mathrm{equal}.\right] \\ \Rightarrow ∆\mathrm{OQP} \cong ∆\mathrm{ORP} \left[\mathrm{by} \mathrm{SAS} \mathrm{property} \mathrm{of} \mathrm{congruence}\right] \\ \Rightarrow \angle \mathrm{QPO} =\angle \mathrm{OPR} =35° \left[\mathrm{by} \mathrm{CPCT}\right] \\ \Rightarrow \angle \mathrm{QPR} =\angle \mathrm{QPO} +\angle \mathrm{OPR} = 35° + 35° = 70° \end{gathered}$$

Regards,