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 – 30⁰  = 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 tangent PA is 6cm and the length of radius OA is 2√3 cm.