[ Law of Sines ]

Find the hypotenuse of a right triangle with an acute angle of $30^{\circ}$, given that the bisector drawn from the vertex of the right angle is equal to $a$.

#
Use the Law of Sines.

## Solution

Let $CD$ be the bisector of the right triangle $ABC$, drawn from the vertex $C$ of the right angle, $\angle A = 30^{\circ}, CD = a$. From triangle $CBD$, by the Law of Sines, we have

$$
\frac{BD}{\sin \angle BCD} = \frac{CD}{\sin \angle B}
$$

from which

$$
BD = \frac{CD \sin \angle BCD}{\sin \angle B} = \frac{a \sin 45^{\circ}}{\sin 60^{\circ}} = \frac{a \sqrt{2}}{\sqrt{3}}
$$

Similarly, from triangle $CAD$, we find that $AD = a \sqrt{2}$. Therefore,

$$
AB = BD + AD = \frac{a \sqrt{2}}{\sqrt{3}} + a \sqrt{2} = \frac{a(\sqrt{6} + 3 \sqrt{2})}{3}.
$$

## Answer

$\frac{a(\sqrt{6} + 3 \sqrt{2})}{3}$.