From an external point P, two tangents PA and PB are

Solution

Let the circle have center O and radius 5 cm. From an external point P, tangents PA and PB touch the circle at A and B. Given ∠APB = 60°. We must find: OP = distance from O to P PA = length of the tangent from P to the circle Use right angles at points of tangency OA and OB are radii drawn to the points of tangency A and B. A radius is perpendicular to the tangent at the point of contact, so: ∠OAP = 90° and ∠OBP = 90°. So quadrilateral OAPB has: ∠OAP = 90°, ∠OBP = 90°, ∠APB = 60°, and ∠AOB is unknown. Find the central angle ∠AOB Sum of interior angles of a quadrilateral is 360°: ∠OAP + ∠OBP + ∠APB + ∠AOB = 360° 90° + 90° + 60° + ∠AOB = 360° 240° + ∠AOB = 360° ∠AOB = 120°. So the angle between radii OA and OB is 120°. Use symmetry to find angles in triangle OAP Tangents from an external point are equal: PA = PB. Also OA = OB (both are radii). Thus triangles OAP and OBP are congruent (by RHS or HL), so OP is the line of symmetry. Therefore, OP bisects both ∠APB and ∠AOB: ∠APO = ∠BPO = 60° / 2 = 30° ∠AOP = ∠BOP = 120° / 2 = 60°. Now consider triangle OAP. Its angles are: ∠OAP = 90° (radius ⟂ tangent) ∠AOP = 60° ∠APO = 30°. So triangle OAP is a 30°–60°–90° right triangle, with right angle at A. Use 30°–60°–90° triangle ratios In a 30°–60°–90° triangle: Side opposite 30° = x Side opposite 60° = x√3 Side opposite 90° (hypotenuse) = 2x. In triangle OAP: Angle at P is 30°, so side opposite 30° is OA. OA is the radius, so OA = 5 cm. Thus: OA = x = 5 cm Then: OP (opposite 90°) = 2x = 2 × 5 = 10 cm PA (opposite 60°) = x√3 = 5√3 cm. Answer OP = 10 cm PA = 5√3 cm