<s>[INST] What's the measure of angle $PCB$ in the figure below? 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} $
...?
 [/INST] 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$
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