Answer
$\cos ^{-1}\dfrac{1}{\sqrt 6}\approx 66^\circ$
Work Step by Step
As we have from the given curves $r_1(t)=\lt t,t^2,t^3 \gt $ and $r_2(t)= \lt \sin t,\sin 2t,t \gt $
Tangent vector $r'(t)$ can be found by taking the derivative of $r_1(t)$ and $r_2(t)$ with respect to $t$.
This yields,$r_1'(t)=\lt 1,2t,3t^2 \gt $ ,$r_2'(t)= \lt \cos t,2 \cos t,1 \gt $
$r_1'(0)=\lt 1,0,0 \gt $ and $r_2'(t)= \lt 1,2,1 \gt $
Consider $\theta$ be the angle between the two curves.
Thus, $\cos \theta=\dfrac{\lt 1,0,0 \gt \cdot\lt 1,2,1 \gt}{\sqrt {1+0+0}\sqrt {1+4+1}}$
or, $ \theta=\cos ^{-1}[\dfrac{\lt 1,0,0 \gt \cdot\lt 1,2,1 \gt}{\sqrt {1+0+0}\sqrt {1+4+1}}]$
or, $=\cos ^{-1}\dfrac{1}{\sqrt 6}\approx 66^\circ$
