In the quadrilateral $A B C D$, $A B=1, B C=2, C D=\sqrt{3}, \angle A B C=120^{\circ}$, and $\angle B C D=90^{\circ}$. What is the exact value of the length of side $A D$?
I. solution. Place the quadrilateral $ABCD$ in the Cartesian coordinate system so that the segment $CD$ lies on the $x$-axis, and point $C$ is at the origin. Then the coordinates of points $B$ and $D$ are: $B(0; 2), D(\sqrt{3}; 0)$. The projection of point $A$ on the $y$-axis is $A'$. Since $\angle ABA' = 60^\circ$, from the triangle $ABA'$, we have $AA' = \sin 60^\circ = \frac{\sqrt{3}}{2}, BA' = \cos 60^\circ = \frac{1}{2}$, and thus the coordinates of point $A$ are: $A\left(\frac{\sqrt{3}}{2}; \frac{5}{2}\right)$. The length of segment $AD$ is:

$$
AD = \sqrt{\left(\frac{\sqrt{3}}{2} - \sqrt{3}\right)^2 + \left(\frac{5}{2}\right)^2} = \sqrt{\frac{28}{4}} = \sqrt{7}
$$

István Kocsis (Miskolc, Földes F. Gymnasium, 11th grade)

![](https://cdn.mathpix.com/cropped/2024_05_02_a673c62950a37dc2cbf8g-1.jpg?height=374&width=282&top_left_y=672&top_left_x=911)

II. solution. Using the cosine rule in triangle $ABC$, we can determine the length of diagonal $AC$:

$$
AC^2 = 1^2 + 2^2 - 2 \cdot 1 \cdot 2 \cdot \cos 120^\circ
$$

From this, $AC = \sqrt{7}$. In the right triangle $BCD$, $DB^2 = 2^2 + (\sqrt{3})^2$, so $DB = \sqrt{7}$. Let the angle $\angle BCA$ be $\alpha$. Using the cosine rule in triangle $CAB$, we get:

$$
\cos \alpha = \frac{(\sqrt{7})^2 + 2^2 - 1}{2 \cdot 2 \sqrt{7}} = \frac{5 \sqrt{7}}{14}
$$

Finally, using the cosine rule in triangle $ADC$, we find the length of the desired side $AD$:

$$
AD^2 = (\sqrt{7})^2 + (\sqrt{3})^2 - 2 \cdot \sqrt{7} \cdot \sqrt{3} \cdot \cos (90^\circ - \alpha)
$$

We know that $\cos (90^\circ - \alpha) = \sin \alpha$. Using the Pythagorean identity, we can calculate $\sin \alpha$ from $\cos \alpha$, given that $\alpha < 90^\circ$:

$$
\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - \left(\frac{5 \sqrt{7}}{14}\right)^2} = \sqrt{\frac{21}{196}} = \frac{\sqrt{3} \cdot \sqrt{7}}{14}
$$

Substituting this into (1) and simplifying, we get $AD = \sqrt{7}$.

III. solution. Extend sides $AB$ and $CD$, and let the intersection of the resulting rays be $E$.

![](https://cdn.mathpix.com/cropped/2024_05_02_a673c62950a37dc2cbf8g-1.jpg?height=355&width=575&top_left_y=2023&top_left_x=772)

In the right triangle $BCE$, $\angle CBE = 60^\circ$, so $\angle BEC = 30^\circ$, $BE = 2BC = 4$, and $CE = \sqrt{3}BC = 2\sqrt{3}$. In triangle $AEB$, $AE = 5$, $DE = 3\sqrt{3}$, and $\angle AED = 30^\circ$. Applying the cosine rule:

$$
AD^2 = AE^2 + DE^2 - 2AE \cdot DE \cdot \cos \angle AED = 25 + 27 - 30\sqrt{3} \cdot \frac{\sqrt{3}}{2} = 7,
$$

so $AD = \sqrt{7}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_a673c62950a37dc2cbf8g-2.jpg?height=412&width=331&top_left_y=130&top_left_x=889)