In a triangle, two angles $\beta$ and $\gamma$ and the radius $R$ of the circumscribed circle are given. Find the radius of the inscribed circle.

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Let in triangle $ABC$, $\angle B = \beta$, $\angle C = \gamma$, $r$ - the radius, $O$ - the center of the inscribed circle, $K$ - the point of tangency of this circle with side $AB$.

Method 1. $AK = \frac{1}{2}(AB + AC - BC) = R(\sin \beta + \sin \gamma - \sin (\beta + \gamma)) = R(\sin \beta (1 - \cos \gamma) + \sin \gamma (1 - \cos \beta)) = 4R \sin \beta / 2 \sin \gamma / 2 \sin (\beta + \gamma / 2)$, and $r = AK \operatorname{ctg} (\beta + \gamma / 2) = 4R \sin \beta / 2 \sin \gamma / 2 \cos (\beta + \gamma / 2)$.

Method 2. $OB \cdot OC \sin \angle BOC = 2S_{BOC} = rBC$, that is, $\frac{r^2}{\sin \frac{\beta}{2} \sin \frac{\gamma}{2}} = 4rR \cos \frac{\beta + \gamma}{2}$, from which $r = 4R \sin \beta / 2 \sin \gamma / 2 \cos (\beta + \gamma / 2)$.

## Answer

$4R \sin \beta / 2 \sin \gamma / 2 \cos (\beta + \gamma / 2)$.