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  • In Fig. 10.9, \angleAOB = 90º and \angleABC = 30º, then \angleCAO is equal to: (A) 30º (B) 45º (C) 90º (D) 60º

In Fig. 10.9, \angleAOB = 90º and \angleABC = 30º, then \angleCAO is equal to:

(A) 30º

(B) 45º

(C) 90º

(D) 60º

Answers (1)

(D)

Solution:

In AOB,

\angleOAB +\angleABO + \angleBOA = 180°      … (i)                           (angle sum property of Triangle)

OA = OB = radius

Angles opposite to equal sides are equal

\angleOAB = \angleABO

Equation (i) becomes

\angleOAB + \angleOAB + 90° = 180°

2\angleOAB = 180° – 90°

\angleOAB  = 45° …(ii)

In \triangleACB,

\angleACB + \angleCBA + \angleCAB = 180°   (angle sum property of Triangle)

\angle ACB = \frac{1}{2} \angle BOA = 45^{\circ}  

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

\therefore  45° + 30° + \angleCAB = 180°

\angleCAB = 180° – 75° = 105°

\angleCAO + \angleOAB = 105°

\angleCAO + 45° = 105°

\angleCAO = 105° – 45° = 60°

Therefore option (D) is correct.

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