Question -
Answer -
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
∠OCA = ∠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 alternate segment are equal]
In ∆PBC, ∠P + ∠C + ∠A= 180°
∠P = 180° – (∠C + ∠A)
∠P = 180° – (110° + 20°)
∠P = 180° – 130° = 50°
In ∆PBC,
∠BPC + ∠PCB + ∠PBC = 180°
[sum of all interior angles of any triangle is 180°]
=> 50° + 20° + ∠PBC = 180°
=> ∠PBC = 180° – 70°
∠PBC = 110°
Since, ∆PB is a straight line.
∠PBC + ∠CBA = 180°
∠CBA = 180° – 110° = 70°