Answer
$cos^{-1}(\frac{50}{\sqrt {4901}})\approx 44^\circ$
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 $|a|=\sqrt {(a_{1})^{2}+(a_{2})^{2}}$
$a.b=-2\times 5+ 5\times 12=-10+60=50$
$|a|=\sqrt {(-2)^{2}+(5)^{2}}=\sqrt {29}$
$|b|=\sqrt {(5)^{2}+(12)^{2}}=\sqrt {169}=13$
Angle between two vectors is given by
$cos\theta=\frac{a.b}{|a||b|}$
$cos\theta=\frac{50}{ \sqrt {29}\times \sqrt {169}}=\frac{50}{\sqrt {4901}}$
$\theta=cos^{-1}(\frac{50}{\sqrt {4901}})\approx 44^\circ$
Hence, $cos^{-1}(\frac{50}{\sqrt {4901}})\approx 44^\circ$
