A) \[\text{9}0{}^\circ \]
B) \[\text{45}{}^\circ \]
C) \[\text{22}.\text{5}{}^\circ \]
D) \[\text{3}0{}^\circ \]
Correct Answer: B
Solution :
\[\vec{A}=\hat{i}+\hat{j}\]Þ \[|A|=\sqrt{{{1}^{2}}+{{1}^{2}}}=\sqrt{2}\] \[\cos \alpha =\frac{{{A}_{x}}}{|A|}=\frac{1}{\sqrt{2}}=\cos 45{}^\circ \] \ \[\alpha =45{}^\circ \]
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