If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°,
Question:
If tangents PA and PB from a point P to a circle with centre $\mathrm{O}$ are inclined to each other at angle of $80^{\circ}$, then $\angle \mathrm{POA}$ is equal to
(A) 50°
(B) 60°
(C) 70°
(D) 80°
Solution:
In figure,
$\Delta \mathrm{OAP} \cong \Delta \mathrm{OBP}(\mathrm{SSS}$ congruence $)$
$\Rightarrow \angle \mathrm{POA}=\angle \mathrm{POB}$
$=\frac{\mathbf{1}}{\mathbf{2}} \angle \mathrm{AOB}$...(1)
Also $\angle \mathrm{AOB}+\angle \mathrm{APB}=180$
$\Rightarrow \angle \mathrm{AOB}=100^{\circ} \ldots(2)$
Then from (1) and (2)
$\angle \mathrm{POA}=\frac{\mathbf{1}}{\mathbf{2}} \times .100=50^{\circ}$
Hence, the correction option is (A)