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 = 300

We know that the angle between the radius and the tangent is 900 at the point of contact,

therefore ∠ OAP = 900

Hence, ∠ BAP = ∠ OAP – ∠ OAB = 900 – 30= 600

Since AP = BP (being tangents to a circle from same external point) ∠ BAP = ∠ PBA  = 600

Now in Δ ABP, by angle sum property, ∠ BAP + ∠ PBA  + ∠ APB = 1800

Hence, ∠ APB = 1800 – ∠ BAP – ∠ PBA

= 1800 – 600 – 600 = 600 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 = 600

Therefore ∠OPA = x 600 = 300

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.

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