3. In triangle $A B C$, the angle $\alpha$ is $30^{\circ}$, side $a$ is $4 \mathrm{~cm}$ long, and side $c$ is twice as long as the median to side $c$. Accurately calculate the lengths of the sides of triangle $A B C$. Draw a sketch.
3. Let point $D$ be the midpoint of side $c$. Triangle $C A D$ is isosceles, so $\angle A C D = 30^{\circ}$ and $\angle C D A = 120^{\circ}$. Therefore, $\angle B D C = 60^{\circ}$. Since triangle $B C D$ is also isosceles, $\angle C B D = \angle D C B = 60^{\circ}$. This means that triangle $A B C$ is a right triangle ($\gamma = 90^{\circ}$). Using trigonometric functions, we can calculate the lengths of sides $b$ and $c$, for example, $\tan 30^{\circ} = \frac{a}{b}$ and $\sin 30^{\circ} = \frac{a}{c}$. We calculate $b = \frac{a}{\tan 30^{\circ}} = 4 \sqrt{3} \text{ cm}$ and $c = \frac{a}{\sin 30^{\circ}} = 8 \text{ cm}$.

![](https://cdn.mathpix.com/cropped/2024_06_07_97c791425a4f6cf5fca5g-09.jpg?height=334&width=606&top_left_y=201&top_left_x=725)
Correct sketch ..... 1 point
Considering that triangles $C A D$ and $B C D$ are isosceles ..... 1 point
Considering that triangle $A B C$ is a right triangle ..... 1 point
Using the relationship $\tan 30^{\circ} = \frac{a}{b}$ ..... $1^{*}$ point
Calculating $c = 8 \text{ cm}$ ..... 1 point
Using the relationship $\sin 30^{\circ} = \frac{a}{c}$ ..... $1^{*}$ point
Calculating $b = 4 \sqrt{3} \text{ cm}$ ..... 1 point
NOTE: If the units are missing in the result, deduct 1 point.