Answer
$cos^{-1}(\frac{1}{\sqrt 5})\approx 63^\circ$
Work Step by Step
The dot product of $a=a_{1}i+a_{2}j$ and $b=b_{1}i+b_{2}j$ is defined as $a.b=a_{1}\times b_{1}+a_{2}\times b_{2}$
The magnitude of a vector is given as $|a|=\sqrt {(a_{1})^{2}+(a_{2})^{2}}$
$a.b=4\times 2+ 3\times (-1)=8-3=5$
$|a|=\sqrt {(4)^{2}+(3)^{2}}=5$
$|b|=\sqrt {(2)^{2}+(-1)^{2}}=\sqrt 5$
Angle between two vectors is given by
$cos\theta=\frac{a.b}{|a||b|}$
$cos\theta=\frac{5}{5\times \sqrt 5}=\frac{1}{\sqrt 5}$
$\theta=cos^{-1}(\frac{1}{\sqrt 5})\approx 63^\circ$
Hence, $cos^{-1}(\frac{1}{\sqrt 5})\approx 63^\circ$
