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