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

Solution

ATQ,

Key facts: OA ⟂ PA and OB ⟂ PB (radii to tangents). Triangles OAP and OBP are congruent right triangles. The angle between radii OA and OB at the centre is ∠AOB = 180° − ∠APB = 180° − 60° = 120°. Length of chord AB: AB = 2R sin(∠AOB/2) = 2 × 5 × sin60° = 10 × (√3/2) = 5√3 cm Also, AB is the base of isosceles triangle APB with PA = PB. In right triangle OAP, let OP = d. Then OA = 5, PA is tangent. Using geometry of two tangents, AB = 2√(OP² − r²) = 2√(d² − 25) So 2√(d² − 25) = 5√3 ⇒ √(d² − 25) = (5√3)/2 Square both sides: d² − 25 = (25 × 3)/4 = 75/4 d² = 25 + 75/4 = (100/4 + 75/4) = 175/4 d = √(175/4) = (√175)/2 = (5√7)/2 cm