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  • In Fig. 10.8, BC is a diameter of the circle and \angleBAO = 60º. Then \angleADC is equal to: Fig. 10.8 (A) 30º (B) 45º (C) 60º (D) 120º

In Fig. 10.8, BC is a diameter of the circle and \angleBAO = 60º. Then \angleADC is equal to:

Fig. 10.8

(A) 30º

(B) 45º

(C) 60º

(D) 120º

Answers (1)

(C)

Solution:

In \triangleAOB, AO = OB    (Radius)

\angleABO = \angleBAO         [Angle opposite to equal sides are equal]

\angleABO = 60° [\therefore \angleBAO = 60° Given]

In \triangleAOB,

\angleABO + \angleOAB =\angleAOC (exterior angle is equal to sum of interior opposite angles)

 60° + 60° = 120° =\angleAOC

\angle ADC =\frac{1}{2} \angle AOC

(Angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle)

\angle ADC =\frac{1}{2} \left ( 120^{\circ} \right )

\angleADC = 60°

Therefore option (C) is correct.

Posted by

infoexpert24

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