# If the magnitudes of vectors $\overrightarrow{A},\overrightarrow{B},\overrightarrow{C}$ are 3,4 and $\theta$ units respectively and if $\overrightarrow{A}+\overrightarrow{B}=\overrightarrow{C},$ then the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is

$\begin{array}{1 1}(A)\;\large\frac{\pi}{4} \\(B)\;\large\frac{\pi}{2} \\(C)\;\tan^{-1}(0.75)\\(D)\;\cos^{-1}(0.6) \end{array}$

Let the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ be $\theta$ . Then
$c^2=A62+B^2 +2AB \cos \theta$
$5^2 =3^2 +4^2 +(2)(3)(4) \cos \theta$
Solving $\theta=90^{\circ}$ .
So, the triangle representing vectors $\overrightarrow{A},\overrightarrow{B}$ and $\overrightarrow{C}$ is a right-angled triangle with $\overrightarrow{C}$ as the hypotenuse. Hence
$\cos \beta = \large\frac{ |\overrightarrow{A}|}{|\overrightarrow{C}|}$
$\qquad= \large\frac{3}{5}$
$\qquad= 0.6$
Hence D is the correct answer.