$[$ Area of a triangle (through semiperimeter and radius of the inscribed or exscribed circle)

In the triangular pyramid $S A B C$, the height $S O$ passes through the point $O$ - the center of the circle inscribed in the base $A B C$ of the pyramid. It is known that $\angle S A C=60^{\circ}, \angle S C A=45^{\circ}$, and the ratio of the area of triangle $A O B$ to the area of triangle $A B C$ is $\frac{1}{2+\sqrt{3}}$. Find the angle $B S C$.

#
Let the circle inscribed in triangle $ABC$ touch the sides $AB, AC$, and $BC$ at points $K, L$, and $M$ respectively. Right triangles $OKS, OLS$, and $OMC$ are equal by two legs, so $SK = SL = SM$. Since $OK \perp AB$, by the theorem of three perpendiculars, $SK \perp AB$. Similarly, $SL \perp AC$ and $SM \perp BC$. Let $SK = SL = SM = h$. Then

$$
AK = AL = SL \operatorname{ctg} 60^{\circ} = \frac{h}{\sqrt{3}}, CM = CL = SL \operatorname{ctg} 45^{\circ} = h
$$

Let $OK = OL = OM = r, BK = BM = x$. Then

$$
\begin{gathered}
S_{\triangle AOB} = \frac{1}{2} AB \cdot OK = \frac{1}{2} (AK + KB) \cdot OK = \frac{1}{2} \left(\frac{h}{\sqrt{3}} + x\right) r \\
S_{\triangle ABC} = \frac{1}{2} (AB + AC + BC) \cdot OK = \frac{1}{2} \left(\frac{h}{\sqrt{3}} + 2h + 2x\right) r = \\
\left(\frac{h}{\sqrt{3}} + h + x\right) r
\end{gathered}
$$

By the condition $\frac{S_{\triangle ABC}}{S_{\triangle AOC}} = 2 + \sqrt{3}$, so

$$
\frac{2 \left(\frac{h}{\sqrt{3}} + h + x\right)}{\sqrt{3} + x} = 2 + \sqrt{3}
$$

from which $BK = BM = x = \frac{h}{\sqrt{3}}$. From the right triangle $BMS$, we find that

$$
\operatorname{tg} \angle BSM = \frac{BM}{SM} = \frac{\frac{h}{\sqrt{3}}}{h} = \frac{1}{\sqrt{3}},
$$

so $\angle BSM = 30^{\circ}$, and since $\angle MCS = \angle LCS = 45^{\circ}$, then $\angle CSM = 45^{\circ}$. Therefore,

$$
\angle BSC = \angle BSM + \angle CSM = 30^{\circ} + 45^{\circ} = 75^{\circ}
$$

## Answer

$75^{\circ}$.

## Problem