12.022. Find the angle of a triangle if it is known that the sides enclosing this angle are 1 and 3, and the bisector of the angle is \(0.75 \sqrt{3}\).
## Solution.

In $\triangle A B C$ we have: $A C=1, B C=3, \angle A C C_{1}=\angle B C C_{1}, C C_{1}=0.75 \sqrt{3}$; we need to find $\angle A C B$ (Fig. 12.25).

We use the formula $l_{c}=\frac{2 a b \cos (C / 2)}{a+b}$ (see "Some relations between elements of figures", p. 713). Expressing $\cos \frac{C}{2}$ from this, we get

$$
\cos \frac{C}{2}=\frac{l_{c}(a+b)}{2 a b}
$$

or, after substituting the values $a=1, b=3, l_{c}=0.75 \sqrt{3}$, we find

$$
\cos \frac{C}{2}=\frac{3 \sqrt{3} \cdot 4}{4 \cdot 2 \cdot 1 \cdot 3}=\frac{\sqrt{3}}{2}
$$

Therefore, $\frac{C}{2}=30^{\circ}$, from which $C=60^{\circ}$.

Answer: $60^{\circ}$.