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  • In Fig. 10.6, if \angle OAB = 40^{\circ} , then \angle ACB is equal to: Fig. 10.6 (A) 50º (B) 40º (C) 60º (D) 70°

In Fig. 10.6, if \angle OAB = 40^{\circ} , then \angle ACB is equal to:

Fig. 10.6

(A) 50º

(B) 40º

(C) 60º

(D) 70°

Answers (1)

(A) 50º

Solution:

AO = OB (Radius of circle)

OBA = \angle OAB = 40^{\circ}  (angles opposite to equal sides in a triangle are equal)

In \triangle OAB

40^{\circ} + \angle AOB + 40 = 180^{\circ} (angle sum property of a triangle)

\angle AOB = 100 ^{\circ}

We know that

2 \angle ACB = \angle AOB  (Angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle)
\angle ACB =\frac{100}{2} \\ \angle ACB = 50^{\circ}

Hence, option A is the correct answer.

Posted by

infoexpert24

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