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  • In Fig.10.5, if AOB is a diameter of the circle and AC = BC, then \angle CAB is equal to: (A) 30º (B) 60º (C) 90º (D) 45º

In Fig.10.5, if AOB is a diameter of the circle and AC = BC, then \angle CAB is equal to:

(A) 30º

(B) 60º

(C) 90º

(D) 45º

Answers (1)

(D)

Solution:

We know that diameter subtends a right angle to the circle.

\therefore \angle BCA = 90^{\circ} \: \: \: \cdots (i)

Now,

AC = BC

\angle ABC =\angle CAB \: \: \: \cdots (ii)        (Angles opposite to equal sides are equal in a triangle)

In \triangle ABC

\angle CAB +\angle ABC +\angle BCA = 180^{\circ}      (By angle sum property of a triangle)
\\\angle CAB +\angle CAB +\angle 90^{\circ} = 180^{\circ} \\ 2\angle CAB = 180^{\circ}-\angle 90^{\circ} \\ \angle CAB = \frac{90^{\circ}}{2}\\ \angle CAB = 45^{\circ}

Therefore option (D) is correct.

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infoexpert24

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