In triangle $ABC$, $\angle CBA = 45^{\circ}$. On side $BC$, there is a point $P$ such that $BP:PC = 1:2$, and $\angle CPA = 60^{\circ}$. Determine $\angle BAC$ without using trigonometric functions.
Let the orthogonal projection of point $C$ onto line $AP$ be denoted by $D$. This cannot coincide with $P$, and it is certainly within the segment $PA$ due to $\angle CPD = 60^\circ$. In the right triangle $CPD$, $\angle PCD = 30^\circ$. Reflect point $P$ over $D$, and let the reflection be $P'$. In the equilateral triangle $CPP'$, $PD = \frac{1}{2} PP' = \frac{1}{2} PC = PB$. Thus, triangle $PDB$ is isosceles and

$$
\angle PBD = \angle BDP = 30^\circ
$$

from which

$$
\angle DBA = 15^\circ,
$$

and

$$
\angle BAP = 180^\circ - (120^\circ + 45^\circ) = 15^\circ
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_d33dc7a275f62522dc65g-1.jpg?height=545&width=629&top_left_y=627&top_left_x=737)

Triangle $ABD$ is also isosceles, with $AD = DB$, and the latter is equal to $CD$ from triangle $BDC$. Therefore, triangle $ACD$ is isosceles and right-angled, so $\angle ACD = 45^\circ$. Thus, for the desired angle $\angle ACB$:

$$
\angle ACB = \angle ACD + \angle DCB = 45^\circ + 30^\circ = 75^\circ.
$$