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प्रश्न
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.
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उत्तर
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
