For reference: In the interior of a triangle ABC, a point $P$ is marked in such a way that: $PC=BC$ and the measure of the angle $PAB$ is equal to the measure of the angle $PAC$ which is $17°$. calculate the measure of angle $PCB$, if the measure of angle $B=107^o$ (Answer:$26^o$)
My progress
$\triangle ABC: \angle C = 180-107-34 = 124+\theta\\ \angle CBP=\angle CPB=90^o - \frac{\theta}{2}\\ \triangle APC: \angle APC = 124^o+\theta\\ \triangle ABP: \angle BPA = 146-\frac{\theta}{2} $
...?
On
Applying Trigonometric form of Ceva's theorem,
$ \displaystyle \sin \angle PAC \cdot \sin \angle PCB \cdot \sin \angle PBA$ $$= \sin \angle ACP \cdot \sin \angle CBP \cdot \sin \angle BAP $$
i.e. $~ \displaystyle \sin 17^\circ \cdot \sin \theta \cdot \sin \left(17^\circ + \frac {\theta}{2}\right)$ $$= \sin (39^\circ - \theta) \cdot \sin \left(90^\circ - \frac {\theta}{2}\right) \cdot \sin 17^\circ$$
$~ \displaystyle 2 \sin \frac {\theta}{2} \cdot \sin \left(17^\circ + \frac {\theta}{2}\right) = \sin (39^\circ - \theta)$
$\cos 17^\circ - \cos (17^\circ + \theta) = \cos (51^\circ + \theta)$
$\displaystyle \cos 17^\circ = \cos ((34^\circ + \theta) - 17^\circ) + \cos ((34^\circ + \theta) + 17^\circ)$
$\displaystyle \cos 17^\circ = 2 \cos (34^\circ + \theta) \cos 17^\circ$
$\cos (34^\circ + \theta) = \frac 12 = \cos 60^\circ$
$ \therefore \theta = 26^\circ$
No trigonometric function is used in this answer.
As shown in the figure above, we start by constructing $\triangle APD$ such that $\triangle APD\cong\triangle APC$.
Since $\triangle ACD$ is an isosceles, from $\angle DAC=34^\circ$ we have $$\angle ADC=\angle ACD=73^\circ$$ It's given that $\angle ACB=39^\circ$, so $$\angle BCD=\angle ACD-\angle ACB=34^\circ$$
Here you might already know where this is going. With $\angle BCD=34^\circ$, we obtain $$\angle CBD=180^\circ-\angle BCD-\angle BDC=73^\circ$$ and it quickly follows that $$BC=CD$$ Now, with $BC=PC$ and $PD=PC$, we have an important conclusion, which is $$PC=CD=PD$$ Hence $\triangle PCD$ is an equilateral triangle, and $\angle PCD=60^\circ$. This implies $$\angle ACP=\angle ACD-\angle PCD=13^\circ$$ and finally, $$\angle PCB=\angle ACB-\angle ACP=\theta = 26^\circ$$ Hope this helps.