Two tangents AP and AQ are drawn to a circle with centre O from an external point A, where P and Q are points on the circle. If AP = 12 cm and ∠PAQ=60°, then the length of chord PQ is:

12 cm

We know that,

sin A = Perpendicular/Hypotenuse

tan A = Perpendicular/Base

We have,

AP = 12 cm

∠PAQ =  60°

In OPAQ,

∠OPA + ∠OQA + ∠PAQ + ∠POQ = 360°

= 90° + 90° + 60° + ∠POQ = 360° - 240°

= ∠POQ = 120°

= ∠POA = 120°/2

= ∠POA = 60°

In ΔOPA

tan 60 = PA/OP

= √3 = 12/OP

= OP = 12/√3

= OP = 4√3

In ΔOPM

sin 60 = PM/OP

= √3/2 = PM/4√3

= PM = (√3/2) 4√3

So, PM = 6

= PQ = 2PM

= PQ = 2 × 6

= PQ = 12 cm