In the triangles below, find the lengths indicated by letters.

We have a right-angled triangle with the side opposite to $\angle C$ ($AB = 2$) and the angle $\angle C = 45^\circ$. We need to find the hypotenuse $BC$.

The sine of an angle in a right-angled triangle is the ratio of the length of the opposite side to the length of the hypotenuse.

Substitute the given values:

We know that $\sin 45^\circ = \frac{\sqrt{2}}{2}$.

Multiply both sides by $2 \cdot BC$:

Divide by $\sqrt{2}$:

Rationalize the denominator by multiplying the numerator and denominator by $\sqrt{2}$:

The value of BC is $\boxed{\mathbf{2\sqrt{2}}}$.

We have a right-angled triangle with the side opposite to $\angle Q$ ($PR = 7$) and the angle $\angle Q = 75^\circ$. We need to find the adjacent side $PQ = a$ and the hypotenuse $QR = c$.

The sum of angles in a triangle is $180^\circ$.

For $\angle Q$, $PR$ is the opposite side and $PQ$ is the adjacent side.

Substitute the given values:

We know that $\tan 75^\circ = 2 + \sqrt{3}$.

Rationalize the denominator:

For $\angle Q$, $PR$ is the opposite side and $QR$ is the hypotenuse.

Substitute the given values:

We know that $\sin 75^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}$.

Rationalize the denominator:

The values are $\boxed{\mathbf{a = 7(2 - \sqrt{3})}}$ and $\boxed{\mathbf{c = 7(\sqrt{6} - \sqrt{2})}}$.