Question

In Fig . 10.69, the tangent at a point C of a circle and a diameter AB when extended intersect at P . If $\angle$PCA =110⁰, find $\angle$CBA. [Hint: Join CO.]
figure

Answer 1

$$\begin{gathered} \angle \mathrm{ACB}=90°\left(\mathrm{Angle}\mathrm{inscribed}\mathrm{in}\mathrm{a}\mathrm{semi}-\mathrm{circle}\right) \\ \angle \mathrm{PCO}=90°\left(\mathrm{PC}\mathrm{is}\mathrm{a}\mathrm{tangent}\mathrm{at}\mathrm{C}\right) \end{gathered}$$
$$\begin{gathered} Now,\angle \mathrm{PCA}=\angle \mathrm{PCO}+\angle \mathrm{OCA} \\ \Rightarrow 110°=90°+\angle \mathrm{OCA} \\ \Rightarrow \angle \mathrm{OCA}=20° \end{gathered}$$
Since, OC = OA (radii of the circle)
$$\begin{gathered} \angle \mathrm{OCA}=\angle \mathrm{OAC}=20° \\ In∆ABC, \\ \angle \mathrm{BAC}+\angle \mathrm{ACB}+\angle \mathrm{CBA}=180° \\ \Rightarrow 90°+20°+\angle \mathrm{CBA}=180° \\ \Rightarrow \angle \mathrm{CBA}=70° \end{gathered}$$