## Law of Sines

[ Area of a triangle (using semiperimeter and radius of inscribed or exscribed circle. In a convex quadrilateral $ABCD$, side $AB$ is $\frac{25}{64}$, side $BC$ is $12 \frac{25}{64}$, and side $AD$ is $6 \frac{1}{4}$. It is known that angle $DAB$ is acute, angle $ADC$ is obtuse, and the sine of angle $DAB$ is $\frac{3}{5}$, the cosine of angle $ABC$ is $-\frac{63}{65}$. A circle with center at point $O$ touches sides $BC$, $CD$, and $AD$. Find $OC$.
Let $P$ be the intersection point of the extensions of segments $DA$ and $CB$. Find the sides and angles of triangle $DCP$ (in which the given circle is inscribed) using the Law of Sines.

## Solution

Let $\angle DAB = \alpha, \angle ABC = \beta$. Then

\[
\begin{gathered}
\sin \alpha = \frac{3}{5}, \cos \alpha = \frac{4}{5}, \cos \beta = -\frac{63}{65}, \sin \beta = \frac{16}{65} \\
\sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta = -\frac{5}{13}
\end{gathered}
\]

Thus, $\alpha + \beta > 180^\circ$. Therefore, the lines $AD$ and $BC$ intersect at point $P$, which is separated from points $D$ and $C$ by the line $AB$. Then

\[
\angle APB = \alpha + \beta - 180^\circ
\]

Using the Law of Sines, we find $PB$ from triangle $APB$:

\[
PB = \frac{AB \sin \alpha}{\sin (\alpha + \beta - 180^\circ)} = -\frac{AB \sin \alpha}{\sin (\alpha + \beta)} = \frac{39}{64}.
\]

Therefore, $PC = PB + BC = 13$.

Using the Law of Sines from triangle $DPC$, we find that

\[
\sin \angle PDC = \frac{PC \sin \angle P}{DC} = \frac{4}{5}
\]

Then $\cos \angle PDC = -\frac{3}{5}$.

Using the Law of Sines from triangle $DCP$, we find that

\[
PD = \frac{PC \sin \angle C}{\sin \angle PDC} = \frac{33}{4}
\]

Let $r$ be the radius of the given circle, and $O$ be its center. Then

\[
r = \frac{2 S_{\triangle DPC}}{PC + DC + DP} = \frac{PC \cdot CD \sin \angle C}{PC + DC + DP} = \frac{3}{2}
\]

Let $K$ be the point of tangency of the circle with side $DC$. From the right triangle $OKC$, we find that

\[
OC = \frac{r}{\sin \frac{1}{2} \angle C} = \frac{r}{\sqrt{1 - \cos \frac{1}{2} \angle C}} = \frac{\sqrt{130}}{2}
\]

![](https://cdn.mathpix.com/cropped/2024_05_06_e3b1ddf947c7dd2094eeg-11.jpg?height=386&width=964&top_left_y=1053&top_left_x=547)

Answer

$\frac{\sqrt{130}}{2}$.

Send a comment