Question

10. Two tangents PA and PB are drawn to the circle with centre O such that APB = 120°. Prove that OP = 2 AP.
[CBSE
11. PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T (see Fig.
the length TP.
[CBSE 2011,​

Answer 1

Answer:

10.Given: O is the centre of the circle. PA and PB are tangents drawn to a circle and ∠APB = 120°.

To prove: OP = 2AP

Proof:

In ΔOAP and ΔOBP,

OP = OP (Common)

∠OAP = ∠OBP (90°) (Radius is perpendicular to the tangent at the point of contact)

OA = OB (Radius of the circle)

∴ ΔOAP is congruent to ΔOBP (RHS criterion)

∠OPA = ∠OPB = 120°/2 = 60° (CPCT)

In ΔOAP,

cos∠OPA = cos 60° = AP/OP

Therefore, 1/2 =AP/OP

Thus, OP = 2AP

Hence, proved.