CBSE Class 10 Answered

Question: In the figure, PQ is a tangent from an external point P to a circle with centre O and OP cuts the circle at T and, QOR is a diameter. If ∠ POR = 130° and S is a point on the circle, find ∠ 1 + ∠ 2,

Given: ∠POR = 130° Given

∠POQ = 50° (since ∠POR and ∠POQ are supplementary)

∠PQO = 90° (∠ formed between the tangent and radius)

Consider triangle POQ, sum of all angles must be equal to 180°

∠POQ + ∠OPQ + ∠PQO = 180°

50° + ∠OPQ + 90° =180°

∠OPQ = 40°

∠OPQ = 1/2 ∠TOR  ... (Angle made by an arc at centre is double the angle made by the same arc at any point on the circle) 

∠OPQ = 1/2 ∠POR = 1/2 x 130° = 65°

∠2 = 65° and ∠1 = 40°

Hence, ∠2 + ∠1 = 105°