Answer
$\theta=cos^{-1}(\frac{5}{ \sqrt {35}\times \sqrt {29}})\approx 80.97^\circ\approx 81^\circ$
Work Step by Step
The dot product of $a=a_{1}i+a_{2}j+a_{3}k$ and $b=b_{1}i+b_{2}j+b_{3}k$ is defined as $a.b=a_{1}\times b_{1}+a_{2}\times b_{2}+a_{3}\times b_{3}$
The magnitude of a vector is given as $|a|=\sqrt {(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}$
$a.b=-2\times 3+ 4\times (-1)+ 3 \times 5=-6-4+15=5$
$|a|=\sqrt {(3)^{2}+(-1)^{2}+(5)^{2}}=\sqrt {35}$
$|b|=\sqrt {(-2)^{2}+(4)^{2}+(3)^{2}}=\sqrt {29}$
Angle between two vectors is given by
$cos\theta=\frac{a.b}{|a||b|}$
$cos\theta=\frac{5}{ \sqrt {35}\times \sqrt {29}}$
$\theta=cos^{-1}(\frac{5}{ \sqrt {35}\times \sqrt {29}})\approx 80.97^\circ\approx 81^\circ$
