A velocity $$\frac{1}{4}$$ m/s is resolved into two components along $$OA$$ and $$OB$$ making angles $$30^\circ$$ and $$45^\circ$$ respectively with the given velocity. Then the component along $$OB$$ is

Let the given velocity be

$$R=\frac14\text{ m/s}$$

Let the components along $$OA$$ and $$OB$$ be $$P$$ and $$Q$$ respectively.

The resultant $$R$$ lies between the directions $$OA$$ and $$OB$$ and makes angles

$$30^\circ$$

and

$$45^\circ$$

with them.

Hence the angle between the component directions is

$$30^\circ+45^\circ=75^\circ$$

Using the triangle of forces (or velocities) and applying Lami's theorem,

$$\frac{P}{\sin45^\circ}=\frac{Q}{\sin30^\circ}=\frac{R}{\sin75^\circ}$$

Therefore,

$$Q=\frac{R\sin30^\circ}{\sin75^\circ}$$

Substituting

$$R=\frac14$$

gives

$$Q=\frac14\cdot\frac{\frac12}{\sin75^\circ}$$

Using

$$\sin75^\circ=\frac{\sqrt6+\sqrt2}{4}$$

we get

$$Q=\frac{1}{2(\sqrt6+\sqrt2)}$$

Rationalising,

$$Q=\frac{\sqrt6-\sqrt2}{8}$$