Given a triangle $A B C$ such that the foot $P$ of the perpendicular from point $C$ to line $A B$ lies inside segment $A B$. From point $P$, perpendiculars $p, q$ are drawn to lines $A C$ and $B C$ (in that order). Let $S$ be the intersection of line $B C$ and line $q$, and $T$ be the intersection of line $A C$ and line $p$. Calculate the measure of angle $A C B$, given that $|\nless A P T|+|\Varangle B P S|=20^{\circ}$.
From the position of point $P$, it follows that the internal angles $\alpha=|\nless B A C|$ and $\beta=|\nless A B C|$ of the given triangle are acute:

![](https://cdn.mathpix.com/cropped/2024_04_17_5cf185c692e63b89ed09g-1.jpg?height=479&width=663&top_left_y=2102&top_left_x=708)

From the right-angled triangles $A P T, B P S$, we obtain the equalities

$$
\alpha=90^{\circ}-|\nless A P T|, \quad \beta=90^{\circ}-|\nless B P S| .
$$

By adding these, we get

$$
\alpha+\beta=180^{\circ}-(|\Varangle A P T|+|\Varangle B P S|)=180^{\circ}-20^{\circ}=160^{\circ} .
$$

Therefore, the angle $\gamma=|\Varangle A C B|$ of the given triangle $A B C$ has the size

$$
\gamma=180^{\circ}-(\alpha+\beta)=180^{\circ}-160^{\circ}=20^{\circ} .
$$

The problem can also be solved using other methods. For example, by comparing the right-angled triangles $A P T$ and $A P C$, and $B P S$ and $B P C$, we obtain the equalities

$$
|\Varangle A C P|=|\Varangle A P T| \text { and }|\Varangle B C P|=|\Varangle B P S| \text {, }
$$

which, when added, yield

$$
|\Varangle A C B|=|\Varangle A C P|+|\Varangle B C P|=|\Varangle A P T|+|\Varangle B P S|=20^{\circ} .
$$

(Another possibility is offered by the quadrilateral $P S C T$, which has right angles at vertices $T, S$ and an angle of $160^{\circ}$ at vertex $P$.)

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