Question

Question 5
In the figure, if AOB is a diameter of the circle and AC = BC, then $\angle CAB$ is equal to

(A) ${30}^{\circ }$
(B) ${60}^{\circ }$
(C) ${90}^{\circ }$
(D) ${45}^{\circ }$

Answer 1

The answer is D.


We know that, diameter subtends a right angle to the circle.
$\begin{array}{ccc}\therefore & \angle BCA={90}^{\circ }& \dots \left(i\right)\\ Given,& AC=BC& \\ \Rightarrow & \angle ABC=\angle CAB& \dots \left(ii\right)\\ & \left[angleoppositetoequalsidesareequal\right]& \\ In\Delta ABC,& \angle CAB+\angle ABC+\angle BCA={180}^{\circ }& \\ & \left[byanglesumpropertyofatriangle\right]& \\ \Rightarrow & \angle CAB+\angle ABC+\angle BCA={180}^{\circ }& [fromEqs.(i\left)and\right(ii\left)\right]\\ \Rightarrow & 2\angle CAB={180}^{\circ }-{90}^{\circ }& \\ \Rightarrow & \angle CAB=\frac{{90}^{\circ }}{2}& \\ \therefore & \angle CAB={45}^{\circ }& \end{array}$