. 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


circle

 

A. 60°
B. 30°
C. 15°
D. 45°
 

 

Best Answer

Answer ||| B

Solution |||

Given – O is the centre of a circle, BOA is its diameter and the tangent at the point P meets BA extended at T. ∠PBO = 30°.

To find – ∠PTA

In ∆ BOP, OB = OP [radius of the same circle]

i.e. the triangle is isosceles

hence, base angles will be equal

i.e. ∠OPB = ∠OBP = 30°

Again, OP is the radius and PT is the tangent and we know that,

tangent is perpendicular to the radius

So, ∠OPT = 90°

In ∆ BPT,

∠PTB = 180° - (30° + 30° + 90°) = 30°

i.e. ∠PTA = 30°

Hence, (B) is the correct option

 

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