Question

In the following figure, BOA is a diameter of a circle and the tangent at a point P meets BA produced at T. If ∠PBO = 30°, then find ∠PTA. Also, show that AP = AT.

Answer 1

Given:

BOA is a diameter of the circle, the tangent at P meets BA produced at T.

∠PBO = 30°.

Step-wise calculation:

1. In triangle OBP, OB = OP (radii)

⇒ Triangle OBP is isosceles

So ∠OPB = ∠PBO = 30°.

2. Sum of angles in ΔOBP

⇒ ∠BOP = 180° – 30° – 30°

= 120°

So the central angle BOP = 120°, hence arc BP = 120°.

3. Angle at A subtending arc BP (inscribed angle) is `∠BAP = 1/2 xx arc  BP = 60^circ`. 

Thus ∠BAP = 60°.

4. BA is a diameter

⇒ ∠APB = 90°   ...(Angle in a semicircle)

5. In ΔAPB, angles are ∠BAP = 60°, ∠APB = 90°.

So ∠PBA = 180° – 60° – 90° = 30°.

6. By the tangent–chord (alternate segment) theorem, the angle between tangent PT and chord PA equals the angle in the opposite arc:

∠TPA = ∠PBA = 30°

7. At point A, TA is the extension of AB, so ∠PAT is the exterior angle supplementary to ∠PAB:

∠PAT = 180° – ∠PAB

= 180° – 60°

= 120°

8. In ΔPAT the three angles are ∠TPA = 30°, ∠PAT = 120°.

So ∠PTA = 180° – 30° – 120° = 30°.