Answer
$cos^{-1}(\frac{8}{ \sqrt {20}\times \sqrt {5}})\approx 36.87^\circ$
or
$cos^{-1}(\frac{8}{ \sqrt {20}\times \sqrt {5}})\approx 37^\circ$
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 $|a|=\sqrt {(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}$
$a.b=8+0+0=8$
$|a|=\sqrt {(4)^{2}+(0)^{2}+(2)^{2}}=\sqrt {20}$
$|b|=\sqrt {(2)^{2}+(-1)^{2}+(0)^{2}}=\sqrt {5}$
Angle between two vectors is given by
$cos\theta=\frac{a.b}{|a||b|}$
$cos\theta=\frac{8}{ \sqrt {20}\times \sqrt {5}}$
$\theta=cos^{-1}(\frac{8}{ \sqrt {20}\times \sqrt {5}})\approx 36.87^\circ$
Hence, $cos^{-1}(\frac{8}{ \sqrt {20}\times \sqrt {5}})\approx 36.87^\circ$
or
$cos^{-1}(\frac{8}{ \sqrt {20}\times \sqrt {5}})\approx 37^\circ$
