Here is the vector triangle. The unknown velocity is $V_b$

**To Find $V_b$ using Components**

Suppose $\hat{i}$ is the unit vector in E direction and $\hat{j}$ is unit vector in S direction. Then

$V_s = 30 \hat{i}$

$V_{bs} = -10 \hat{i} \cos 45^{\circ} + 10 \hat{j} \sin 45^{\circ} = \frac{10}{\sqrt{2}}(-\hat{i}+\hat{j})$

So

$V_b=V_s+V_{bs} = (30-\frac{10}{\sqrt2})\hat{i} +\frac{10}{\sqrt2} \hat{j}$

The question asks for the magnitude of $V_b$ which is

$|V_b| = \sqrt{ (30-\frac{10}{\sqrt2})^2 + (\frac{10}{\sqrt2} )^2}$

**To Find $V_b$ using the Cosine Rule**

$V_b^2 = V_s^2+V_{bs}^2-2V_sV_{bs}\cos45^{\circ}$

$ = 30^2+10^2 - 2\times 30\times 10\times \frac{1}{\sqrt2}$