The angle made by the vector $A=\hat{i}+\hat{j}$ with x- axis is

The correct option is D: ${45}^{\circ }$

Let $\alpha$ is the angle made by the vector A=$\hat{i}+\hat{j}$ with x- axis then:

$A_x=1,A_y=1$

$\overrightarrow{A}=\hat{i}+\hat{j}\Rightarrow |A|=\sqrt{1^2+1^2}=\sqrt{2}$
$cos\alpha =\frac{A_x}{|A|}=\frac{1}{\sqrt{2}}=cos{45}^{\circ }\therefore \alpha ={45}^{\circ }$