12.136. The lateral sides of the trapezoid are equal to $p$ and $q(p<q)$, the larger base is equal to $a$. The angles at the larger base are in the ratio 2:1.

Find the smaller base.
Solution.

Let $B M$ and $C N$ be the heights of trapezoid $A B C D$ (Fig. 12.9a), $A B=p, C D=q, A D=a$.

Then $\angle A: \angle D=2: 1, M N=B C=x$. Let $\angle D=\alpha$. Then $\angle A=2 \alpha$.

In $\triangle A M B\left(\angle A M B=90^{\circ}\right): B M=p \sin 2 \alpha ; A M=p \cos 2 \alpha$.

In $\triangle D N C\left(\angle D N C=90^{\circ}\right): C N=q \sin \alpha ; D N=q \cos \alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0898.jpg?height=462&width=636&top_left_y=72&top_left_x=356)

Fig. 12.96

Since $B M=C N$, then $p \sin 2 \alpha=q \sin \alpha ; 2 p \sin \alpha \cos \alpha=q \sin \alpha ; \cos \alpha=\frac{q}{2 p}$. $A D=A M+M N+N D ; a=p \cos 2 \alpha+\mathrm{x}+q \cos \alpha ;$

\[
\begin{aligned}
& x=a-p \cos 2 \alpha-q \cos \alpha=a-p\left(2 \cos ^{2} \alpha-1\right)-q \cos \alpha= \\
& =a-p\left(2 \cdot \frac{q^{2}}{4 p^{2}}-1\right)-q \cdot \frac{q}{2 p}=a-\frac{q^{2}}{2 p}+p-\frac{q^{2}}{2 p}=\frac{p^{2}+a p-q^{2}}{p}
\end{aligned}
\]

In the case where $90^{\circ}<2 \alpha<180^{\circ}$ (Fig. 12.9b) $A D=N D+M N-A M=$ $q \cos \alpha+x-p \cos \left(180^{\circ}-2 \alpha\right)=q \cos \alpha+x+q \cos 2 \alpha$, and we get the same answer.

Answer: $\frac{p^{2}+a p-q^{2}}{p}$.