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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:$26^o$)

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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} $

...?

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  • $\begingroup$ Simple angle chasing isn't going to be enough to solve this, by the looks of it. You'll want to try some different techniques. My instincts tell me to try applying the Law of Sines, since that behaves well with angle bisectors and isosceles triangles. $\endgroup$ Mar 8, 2022 at 21:17
  • $\begingroup$ Angel C = 39? Calculate $\theta$. $\endgroup$
    – Moti
    Mar 9, 2022 at 2:12

2 Answers 2

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No trigonometric function is used in this answer.

enter image description here

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.

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    $\begingroup$ +1 and I inserted the diagram in the answer $\endgroup$
    – Math Lover
    Mar 9, 2022 at 8:04
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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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  • $\begingroup$ Link to trigonometric form of Ceva's theorem: cut-the-knot.org/triangle/TrigCeva.shtml $\endgroup$
    – Math Lover
    Mar 9, 2022 at 4:58
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    $\begingroup$ thanks for alternative...I didn't know this trigonometric form of the theorem $\endgroup$ Mar 9, 2022 at 12:17
  • $\begingroup$ you are welcome! $\endgroup$
    – Math Lover
    Mar 9, 2022 at 12:23

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