In a convex quadrilateral $A B C D$, the length of side $A D$ is 4, the length of side $C D$ is 7, the cosine of angle $A D C$ is $\frac{1}{2}$, and the sine of angle $B C A$ is $\frac{1}{3}$. Find the side $B C$, given that the circle circumscribed around triangle $A B C$ also passes through point $D$.
Apply the Law of Cosines to triangle $A D C$ and the Law of Sines to triangle $A B C$.

## Solution

Since $\cos \angle A D C=\frac{1}{2}$, then $\angle A D C=60^{\circ}$, and since quadrilateral $A B C D$ is cyclic, $\angle A B C=180^{\circ}-\angle$ $A D C=120^{\circ}$.

By the Law of Cosines

$$
A C=\sqrt{D A^{2}+D C^{2}-2 \cdot D A \cdot D C \cdot \cos \angle A D C}=\sqrt{16+49-2 \cdot 4 \cdot 7 \cdot \frac{1}{2}}=\sqrt{37}
$$

Let $\angle A C B=\alpha$. Note that angle $A C B$ is acute (since angle $A B C$ is obtuse). Therefore,

$$
\cos \angle A C B=\cos \alpha=\sqrt{1-\sin ^{2} \alpha}=\frac{2 \sqrt{2}}{3}
$$

By the Law of Sines

$$
\frac{B C}{\sin \angle B A C}=\frac{A C}{\sin \angle A B C} \text {, or } \frac{B C}{\sin \left(60^{\circ}-\alpha\right)}=\frac{\sqrt{37}}{\frac{\sqrt{3}}{2}} \text {. }
$$

From this, we find that

$$
\begin{aligned}
B C & =\frac{\sqrt{37}}{\frac{\sqrt{3}}{2}} \cdot \sin \left(60^{\circ}-\alpha\right)=\frac{\sqrt{37}}{\frac{\sqrt{3}}{2}} \cdot\left(\sin 60^{\circ} \cos \alpha-\cos 60^{\circ} \sin \alpha\right)= \\
& =\frac{\sqrt{37}}{\frac{\sqrt{3}}{2}} \cdot\left(\frac{\sqrt{3}}{2} \cdot \frac{2 \sqrt{2}}{3}-\frac{1}{2} \cdot \frac{1}{3}\right)=\frac{\sqrt{37}}{3 \sqrt{3}}(\sqrt{24}-1)
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_4c124ad32a76c9faf91dg-36.jpg?height=566&width=600&top_left_y=883&top_left_x=729)

Answer

$\frac{\sqrt{37}}{3 \sqrt{3}}(\sqrt{24}-1)$

Submit a comment