Question

prove that the angle between the two tangents drawn from the external point to the circle in a supplementary of the angle subtended by the line segment joining the point of contact in a circle sector​

Answer 1

Answer:

Step-by-step explanation:

PA and PB are the two tangents drawn from an external point P at the point of contacts A and B on the circle with centre O respectively.

∴ OA ⊥ PA and OB ⊥ PB

[∵ radius of a circle is always ⊥ to the tangent at the point of contact.]

∴ ∠ OAP = ∠ OBP = 90°

we know that –

Sum of all the angles of a quadrilateral = 360°

In quadrilateral OAPB,

∠ OAP + ∠ OBP + ∠ APB + ∠ AOB = 360°

⇒ 180° + ∠ APB + ∠ AOB = 360°

∴ ∠ APB + ∠ AOB = 180°

Hence, the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact to the centre.