**Q) **PA and PB are tangents drawn to a circle of centre O from an external point P. Chord AB makes and angle of 30 with the radius at the point of contact. If length of the chord of 6 cm, find the length of the tangent PA and the length of the radius OA.

**Ans: **

In the given diagram, it is given that: ∠ OAB = 30^{0}

We know that the angle between the radius and the tangent is 90^{0 }at the point of contact,

therefore ∠ OAP = 90^{0}

Hence, ∠ BAP = ∠ OAP – ∠ OAB = 90^{0 }– 30^{0 }= 60^{0}

Since AP = BP (being tangents to a circle from same external point)

∠ BAP = ∠ PBA ^{ }= 60^{0}

Now in Δ ABP, by angle sum property, ∠ BAP + ∠ PBA + ∠ APB = 180^{0}

Hence, ∠ APB = 180^{0} – ∠ BAP – ∠ PBA

= 180^{0} – 60^{0} – 60^{0} = 60^{0}

Δ ABP is an equilateral triangle.

**and AP = BP = AB = 6 cm (given)**

Next, in Δ OAP and Δ OBP:

OA = OB (radius of same circle)

PA = PB (tangent on a circle from same external point)

OP = OP (common side)

Therefore, Δ OAP Δ OBP:

Hence, ∠OPA = ∠OPB

Since, we just calculated above that ∠APB = 60^{0}

Therefore ∠OPA = x 60^{0} = 30^{0}

In Δ OAP, tan 30 =

or

OA = = **2√3 cm**

**Therefore, the length of the chord AB is 6cm and the length of radius OA is 2√3 cm.**