Question

In the given figure, BOA is a diameter of the circle and the tangent at a point P to the circle meets BA extended at T, if angle PBO=30, find angle PTA

Answer 1

join OP and radius is same
∠OPB=30
∠POB=120......sum of triangle
∠POT=60...linear pair
∠TPO=90
then,∠PTO=30.....sum of triangle property
30 is answer.

Answer 2

Given-

1) Angle PBO = 30°

2) BOA is diameter,i.e,[angle BOA=180°]

To Find-

angle PTO

Proof-

Angle PBO=Angle OPB=30° (equal radii)

Angle OPB + Angle OBP + Angle BOP = 180° ( Angle Sum Property)

30° + 30° + Angle BOP =180°

60° + Angle BOP = 180°

Angle BOP= 180°- 60°

Angle BOP = 120°

Now,

BOA is a diameter

Angle BOP + Angle POA =180°(linear pair)

120° + Angle POA = 180°

Angle POA= 180°-120° = 60°      -- (1)

In ∆TOP

Angle POT = 2 Angle PTO ( angle subtended by the centre of the circle is double at any point from the same base)

60° = 2 Angle PTO

60°/2 = Angle PTO

30° = Angle PTO

Proved ☺️