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^∘

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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).

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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∘

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^∘