Task B-4.3. Let $A B C A^{\prime} B^{\prime} C^{\prime}$ be a right regular triangular prism. The length of the base edge $A B C$ is $10 \mathrm{~cm}$, and the height of the prism is $10 \sqrt{2} \mathrm{~cm}$. Point $D$ is the midpoint of edge $\overline{A B}$. Calculate the measure of angle $D A^{\prime} C$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_7c0a9615fb5cd46ea1c5g-19.jpg?height=580&width=422&top_left_y=1122&top_left_x=819)

It is sufficient to calculate the lengths of any two sides of triangle $D A^{\prime} C$.

$|D C|$ is the height of an equilateral triangle, so $|D C|=5 \sqrt{3} \mathrm{~cm}$.

Triangle $D A A^{\prime}$ is a right triangle, so $\left|D A^{\prime}\right|=\sqrt{(10 \sqrt{2})^{2}+5^{2}}=15 \mathrm{~cm}$.

Triangle $C A A^{\prime}$ is a right triangle, so $\left|C A^{\prime}\right|=\sqrt{(10 \sqrt{2})^{2}+10^{2}}=10 \sqrt{3} \mathrm{~cm}$.

(Or $\triangle C D A^{\prime}$ is a right triangle, so $\left|C A^{\prime}\right|=\sqrt{(5 \sqrt{3})^{2}+15^{2}}=10 \sqrt{3} \mathrm{~cm}$.)

(2 points)

We calculate the angle using any of the trigonometric ratios in the right triangle, depending on which two sides we have calculated. For example,

$$
\operatorname{tg} \angle D A^{\prime} C=\frac{|D C|}{\left|D A^{\prime}\right|}=\frac{5 \sqrt{3}}{15}=\frac{\sqrt{3}}{3}
$$

so the desired angle $\angle D A^{\prime} C=30^{\circ}$.