Task B-3.3. The length of the larger base of the trapezoid is $18 \mathrm{~cm}$, and the lengths of the legs of the trapezoid are $b=12 \mathrm{~cm}$ and $d=8 \mathrm{~cm}$. The ratio of the angles at the larger base is $2: 1$. Determine the length of the smaller base of the trapezoid.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_6bd7d8198e5de61c26bbg-12.jpg?height=455&width=854&top_left_y=1613&top_left_x=603)

Let $\alpha$ be the angle $\Varangle E B C$ and apply the Law of Sines to triangle $E B C$:

$$
\frac{8}{\sin \alpha}=\frac{12}{\sin 2 \alpha}
$$

From this, we get

$$
\begin{aligned}
& \frac{\sin \alpha}{2 \sin \alpha \cos \alpha}=\frac{2}{3} \\
& \cos \alpha=\frac{3}{4}
\end{aligned}
$$

Using the Law of Cosines, we have

$$
\begin{aligned}
& 8^{2}=12^{2}+(18-c)^{2}-24(18-c) \cos \alpha \\
& 64=144+(18-c)^{2}-18(18-c)
\end{aligned}
$$

This gives us the equation

$$
c^{2}-18 c+80=0
$$

The solutions are $c=8$ and $c=10$.

In the case $c=10$, triangle $C E B$ is isosceles with sides $8, 8, 12$ and angles $45^{\circ}$, $45^{\circ}, 90^{\circ}$, which means that the Pythagorean theorem $8^{2}+8^{2}=12^{2}$ must hold for its sides. Since this equality does not hold, the only solution is $c=8 \mathrm{~cm}$.

If the Law of Cosines is not used, but the Law of Sines is applied again to triangle $C E B$, we get the equality

$$
\frac{18-c}{\sin \left(180^{\circ}-3 \alpha\right)}=\frac{8}{\sin \alpha}
$$

which, after simplification, gives the solution $c=8$.