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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.
Solution
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
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