The tangent at a point C of a circle and a diameter AB when extended intersect at P. If ∠PCA = 110°, find ∠CBA see figure

#### Solution

Here, AB is a diameter of the circle from point C and a tangent is drawn which meets at a point P.

Join OC, Here, OC is radius.

Since, tangent at any point of a circle is perpendicular to the radius through point of contact circle.

∴ OC ⊥ PC

Now, ∠PCA = 110° ......[Given]

⇒ ∠PCO + ∠OCA = 110°

⇒ 90° + ∠OCA = 110°

⇒ ∠OCA = 20°

∴ OC = OA = Radius of circle

⇒ OC = ∠OAC = 20° .......[Since, two sides are equal. then their opposite angles are equal]

Since, PC is a tangent, so ∠BCP = ∠CAB = 20 ° ......[Angles in a alternative segment are equal]

In ΔPBC, ∠P + ∠C + ∠A = 180°

∠P = 180° – (∠C + ∠A)

= 180° – (110° + 20°)

= 180° – 130° = 50°

In ΔPBC, ∠BPC + ∠PCB + ∠PBC = 180° ......[Sum of all interior angles of any triangle is 180°]

⇒ 50° + 20° + ∠PBC = 180°

⇒ ∠PBC = 180° – 70°

⇒ ∠PBC = 110°

Since, APB is a straight line.

∴ ∠PBC + ∠CBA = 180°

⇒ ∠CBA = 180° – 110° = 70°