## Task B-2.3.

In triangle $A B C$, $|B C|=2 \text{ cm}, |A C|=3 \text{ cm}$, and $\cos \alpha=\frac{7}{8}$, where $\alpha=\angle C A B$.

If the angle between the median and the altitude drawn from vertex $C$ is $\omega$, what is $\cos 2 \omega$?
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_1ebddb66d71be8305afcg-07.jpg?height=439&width=736&top_left_y=523&top_left_x=657)

Let $D$ be the foot of the altitude drawn from vertex $C$. From triangle $A D C$ we calculate

$$
|A D|=b \cdot \cos \alpha=3 \cdot \frac{7}{8}=\frac{21}{8} \text{ cm}
$$

The height can be calculated using the Pythagorean theorem

$$
v=\sqrt{9-\left(\frac{21}{8}\right)^{2}}=\frac{3 \sqrt{15}}{8} \text{ cm}
$$

and we can also calculate $|D B|$

$$
|D B|=\sqrt{4-\left(\frac{3 \sqrt{15}}{8}\right)^{2}}=\frac{11}{8} \text{ cm}
$$

Then

$$
|A B|=|A D|+|D B|=\frac{21}{8}+\frac{11}{8}=4 \text{ cm}
$$

Triangle $C P B$ has two equal sides, so $\angle P C B = \angle C P B = \frac{1}{2}(180^\circ - \beta)$.

Now, $\angle P C D = \frac{1}{2}(180^\circ - \beta) - (90^\circ - \beta) = \frac{\beta}{2}$.

Therefore, $\cos 2\omega = \cos \beta$, and from triangle $C D B$ we get

$$
\cos \beta=\frac{\frac{11}{8}}{2}=\frac{11}{16}
$$