Q. 104.4( 16 Votes )

In figure, if PQR is the tangent to a circle at Q whose center is 0, AB is a chord parallel to PR and BQR = 70°, then AQB is equal to


A. 20°

B. 40°

C. 35°

D. 45°

Answer :


Given : PQR is a tangent to a circle Q with center O and AB is a chord parallel to PR and BQR = 70°


To Find : AQB


Now, OQR = DQR = 90° [ As tangent at any point on the circle is perpendicular to the radius through point of contact]


DQR = QBD + BQR


90 = QBD + 70°


QBD = 20° [1]


As, AB || PR


BDQ + DQR = 180° [ some of co interior angles ]


BDQ + 90° = 180°


BDQ = 90°


And


ADQ + BDQ = 180° [Linear pair]


ADQ + 90° = 180°


ADQ = 90° = BDQ


In AQD and BQD


QD = QD [common]


ADQ = BDQ [Proved Above]


AD = BD [since, perpendicular from center to the chord bisects the chord]


AQD BQD [By Side Angle Side Criterion]


And


AQD = BQD [ Corresponding parts of congruent triangles are equal ]


AQD + BQD = AQB


AQB = 2BQD [As AQD = BQD ]


AQB = 2(20) = 40° [From 1]

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