Problem 11.6. Given a convex quadrilateral $A B C D$. It is known that $\angle C=57^{\circ}, \sin \angle A+$ $\sin \angle B=\sqrt{2}$ and $\cos \angle A+\cos \angle B=2-\sqrt{2}$. How many degrees does angle $D$ measure?
Answer: 168.

Solution. We transform the given sums of trigonometric functions:

\[
\begin{aligned}
\sqrt{2} & =\sin \angle A+\sin \angle B=2 \sin \left(\frac{\angle A+\angle B}{2}\right) \cos \left(\frac{\angle A-\angle B}{2}\right) \\
2-\sqrt{2} & =\cos \angle A+\cos \angle B=2 \cos \left(\frac{\angle A+\angle B}{2}\right) \cos \left(\frac{\angle A-\angle B}{2}\right)
\end{aligned}
\]

Since both expressions are not equal to 0, we divide the first by the second and cancel the common factor \(\cos \left(\frac{\angle A-\angle B}{2}\right)\), which is not equal to 0. We get

\[
\frac{\sqrt{2}}{2-\sqrt{2}}=\operatorname{tg}\left(\frac{\angle A+\angle B}{2}\right)
\]

from which it is not difficult to extract \(\operatorname{tg}\left(\frac{\angle A+\angle B}{2}\right)=\sqrt{2}+1\), by eliminating the irrationality in the denominator.

Applying the double-angle tangent formula \(\operatorname{tg} 2 x=\frac{2 \operatorname{tg} x}{1-\operatorname{tg}^{2} x}\), we obtain

\[
\operatorname{tg}(\angle A+\angle B)=\frac{2(\sqrt{2}+1)}{1-(\sqrt{2}+1)^{2}}=\frac{2 \sqrt{2}+2}{-2 \sqrt{2}-2}=-1
\]

i.e., \(\angle A+\angle B=180^{\circ} \cdot k-45^{\circ}\) for some integer \(k\).

The angles \(\angle A\) and \(\angle B\) are less than \(180^{\circ}\), since the quadrilateral is convex; and since the sum of their cosines is positive, they cannot both be obtuse. Therefore, \(\angle A+\angle B<270^{\circ}\), leaving the only solution \(\angle A+\angle B=135^{\circ}\). Then the angle \(D\) is equal to \(360^{\circ}-57^{\circ}-135^{\circ}=168^{\circ}\).

![](https://cdn.mathpix.com/cropped/2024_05_06_2befe970655743580344g-19.jpg?height=390&width=458&top_left_y=1452&top_left_x=498)

Figure 8: to the remark in the solution of problem 11.6

Remark. Consider the vectors \(\vec{a}=(\cos \angle A, \sin \angle A)\) and \(\vec{b}=(\cos \angle B, \sin \angle B)\) (Figure 8) and their sum, which is equal to \((2-\sqrt{2}, \sqrt{2})\) by the given condition. If we draw the parallelogram corresponding to this sum, it turns out to be a rhombus, and its diagonal is the bisector between \(\vec{a}\) and \(\vec{b}\). This provides a geometric explanation for why the angle between the x-axis and the vector \(\vec{a}+\vec{b}\) is \(\frac{1}{2}(\angle A+\angle B)\), which is why, when dividing one coordinate of the vector by the other in the solution, we obtained \(\operatorname{tg}\left(\frac{\angle A+\angle B}{2}\right)\).