Question

In the figure, O is the center of the circle. Line AQ is a tangent. If OP = 3, m(arc PM) = 120°, then find the length of AP.

Answer 1

Given: line AQ is a tangent.

OP = 3, m(arc PM) = 120°

To find: AP

In the given figure, arc PMQ is a semicircle.

∴ m(arc PMQ) = 180°     ......[Measure of semicircular arc is 180°]

∴ m(arc PM) + m(arc MQ) = 180°    .....[Arc addition property]

∴ 120° + m(arc MQ) = 180°   ......[Given]

∴ m(arc MQ) = 180° – 120°

∴ m(arc MQ) = 60°    .......(i)

∠MPQ = `1/2` m(arc MQ)    .....[Inscribed angle theorem]

∴ ∠MPQ = `1/2 xx 60^circ`     ......[From (i)]

∴ ∠MPQ = 30°

i.e., ∠APQ = 30°    ......(ii) [A – M – P]

In ∆PQA, ∠PQA = 90°    ......[Tangent theorem]

∠ APQ = 30°     ......[From (ii)]

∴ ∠PAQ = 60°    ......[Remaning angle of ∆PQA]

∴ ∆PAQ is 30° – 60° – 90° triangle.

∴ PQ = `sqrt(3)/2` AP    ......[Side opposite to 60°]

∴ (PO + OQ) = `sqrt(3)/2` AP    ......[P – O – Q]

∴ (3 + 3) = `sqrt(3)/2` AP     ......[Radii of same circle and op = 3]

∴ AP = `(6 xx 2)/sqrt(3)`

∴ AP = `(6 xx 2 xx sqrt(3))/(sqrt(3) xx sqrt(3))`  ......[Multiply and divide by `sqrt(3)`]

∴ AP = `(6 xx 2 xx sqrt(3))/3`

∴ AP = `2 xx 2sqrt(3)`

∴ AP = `4sqrt(3)` units