Question

In the figure, tangent at a point C of a circle and a diameter AB when extended intersect at P. If $\angle PCA={110}^{\circ }$, find $\angle CBA$

Answer 1

∠PCA = 110°
PC is the tangent to the circle whose centre is O.
Construction
Join points C and O.

∠BCA = 90° [Since angle in a semi circle is 90°]
Also ∠PCO = 90° [Since radius ⊥ tangent]

From the figure we have,
∠PCA =∠PCO + ∠OCA
i.e. 110° = 90° + ∠OCA
$\therefore$∠OCA =20°

Now in ΔAOC, AO = OC [Radii]
So, ∠OCA = ∠OAC =20°

In ΔABC, we have
∠BCA = 90° & ∠CAB = 20°
$\therefore$∠CBA = 70°