$\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} $

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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.

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