Answer
$\theta =0^\circ$ at $x=0$ and $\theta =8.13^\circ$ at $x=1$
Work Step by Step
$f(x)=x^2$ and $g(x)=x^3$
$x^2=x^3$ $ \implies$ $x=0,1$
$f'(x)=2x$ and $g'(x)=3x^2$
$f'(0)=0$ and $g'(0)=0$
$f'(1)=2$ and $g'(1)=3$
Thus, $a= \lt 1,2\gt$ and $b =\lt 1,3 \gt$
$ \theta =0 ^\circ$ at $x=0$
$ \theta = cos^{-1}\dfrac{a \cdot b}{|a||b|}=cos^{-1}\dfrac{1(1)+2(3)}{\sqrt {5}\sqrt {10}}$
$ \theta = cos^{-1}\dfrac{a \cdot b}{|a||b|}=cos^{-1}\dfrac{7}{\sqrt {50}}\approx 8.13^\circ$ at $x=1$
Hence, $\theta =0^\circ$ at $x=0$ and $\theta =8.13^\circ$ at $x=1$
