Question:
In the given figure, O is the centre of the circle. AB is the tangent to the circle at the point P. If ∠PAO = 30∘ then ∠CPB + ∠ACP is equal to
(a) 60∘
(b) 90∘
(c) 120∘
(d) 150∘
Solution:
We know that a chord passing through the centre is the diameter of the circle.
∵∠DPC = 90∘ (Angle in a semi circle is 90∘)
Now, In △CDP
∠CDP + ∠DCP + ∠DPC = 180∘ [Angle sum property of a triangle]
⇒ ∠CDP + ∠DCP + 90∘ = 180∘
⇒ ∠CDP + ∠DCP = 90∘
By using alternate segment theorem
We have ∠CDP = ∠CPB
∴∠CPB + ∠ACP = 90∘
Hence, the correct answer is option (b).