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# CBSE Class 10 Answered

question no 51
Asked by vs4247325 | 04 Sep, 2020, 04:56: AM

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°

Answered by Renu Varma | 04 Sep, 2020, 01:26: PM

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