Given that, velocity of $\operatorname{car} A$,
$$
\begin{aligned}
& v_A=30 \mathrm{~km} / \mathrm{h} \text { (toward east) } \\
& \Rightarrow \mathbf{v}_A=30 \hat{\mathbf{i}}
\end{aligned}
$$
For car $B, v_B=30 \mathrm{~km} / \mathrm{h}$ (toward North)
$$
\Rightarrow \quad \mathbf{v}_B=30 \hat{\mathbf{j}}
$$


According to question,
Velocity of car $B$ as measured in car $A=$ relative velocity of car $B$ w.r.t car $A$.
$$
\mathbf{v}_{B A}=\mathbf{v}_B-\mathbf{v}_A=30 \hat{\mathbf{j}}-30 \hat{\mathbf{i}}
$$
$\therefore$ Magnitude of $\mathbf{v}_{B A}$
$$
\begin{aligned}
& =\sqrt{v_B^2+v_A^2}=\sqrt{30^2+30^2} \\
& =30 \sqrt{2} \mathrm{~km} / \mathrm{h} \approx 42 \mathrm{~km} / \mathrm{h}
\end{aligned}
$$
Direction of $\mathbf{v}_{B A}$,
$$
\begin{aligned}
\tan \theta & =\frac{v_B}{v_A}=\frac{30}{-30}=-1 \\
\Rightarrow \quad \theta & =135^{\circ}
\end{aligned}
$$
Hence, velocity of car $B$ as measured in car $A$ is $42 \mathrm{~km} / \mathrm{h}$ at angle of $\left(135^{\circ}-90^{\circ}=45^{\circ}\right)$ North of West.