## Task 4 - 180934

In a right-angled triangle $A B C$ with the right angle at $C$, the height dropped from $C$ to the hypotenuse $A B$ divides it in the ratio $1: 3$.

Calculate the size of the interior angles at $A$ and $B$ of the triangle $A B C$!
Let $H$ be the foot of the altitude from $C$ to $A B$, and let $|H B|=3 \cdot|A H|$. Furthermore, let $\alpha:=\angle B A C$ and $\beta:=\angle A B C$.

Then $|A B|=4 \cdot|A H|$, and by the leg theorem, $|B C|=\sqrt{|A H| \cdot|A B|}=2 \cdot|A H|$.

By the definition of the sine in a right triangle, $\sin \alpha=\frac{|B C|}{|A B|}=\frac{1}{2}$, so $\alpha=30^{\circ}$, since $0<\alpha<90^{\circ}$.

From the sum of the interior angles in the triangle, we finally get $\beta=180^{\circ}-90^{\circ}-30^{\circ}=60^{\circ}$.