Question

In the given figure, O is the centre of a circle, BOA is its diameter and the tangent at the point P meets BA extended at T. If ∠PBO = 30^∘ then ∠PTA = ?
(a) 60^∘
(b) 30^∘
(c) 15^∘
(d) 45^∘

Answer 1

We know that a chord passing through the centre is the diameter of the circle.
∵∠BPA = 90^∘ (Angle in a semi circle is 90^∘)
By using alternate segment theorem
We have ∠APT = ∠ABP = 30^∘
Now, In △ABP
∠PBA + ∠BPA + ∠BAP = 180⁰ [Angle sum property of a triangle]
⇒ 30^∘ + 90⁰ + ∠BAP = 180^∘
⇒ ∠BAP = 60^∘
Now, ∠BAP = ∠APT + ∠PTA
⇒ 60^∘ = 30^∘ + ∠PTA
⇒ ∠PTA = 30^∘
Hence, the correct answer is option (b).