In the figure,if AOB is a diameter of the cir | vedclass

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(C) Since $AOB$ is a diameter of the circle,the angle subtended by it at the circumference is $90^{\circ}$.Therefore,$\angle ACB = 90^{\circ}$ (Angle

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(C) Since $AOB$ is a diameter of the circle,the angle subtended by it at the circumference is $90^{\circ}$.Therefore,$\angle ACB = 90^{\circ}$ (Angle in a semi-circle is $90^{\circ}$).Given that $AC = BC$,the triangle $	riangle ABC$ is an isosceles triangle.In an isosceles triangle,angles opposite to equal sides are equal,so $\angle CAB = \angle CBA$.Let $\angle CAB = \angle CBA = x$.In $	riangle ABC$,the sum of angles is $180^{\circ}$:$\angle CAB + \angle CBA + \angle ACB = 180^{\circ}$$x + x + 90^{\circ} = 180^{\circ}$$2x = 90^{\circ}$$x = 45^{\circ}$Thus,$\angle CAB = 45^{\circ}$.

In the figure,if $AOB$ is a diameter of the circle and $AC = BC$,then $\angle CAB$ is equal to (in $^{\circ}$)

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