angle Ceva's theorem

Question

In triangle $ABC$, $P$ is point such that $\angle PAB = 42^{\circ}$, $\angle PBA = 54^{\circ}$, $\angle PAC = 6^{\circ}$, $\angle PBC = 12^{\circ}$. Find $ \angle PCB$.

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I found the $ \angle PCB = 42^{\circ} $ , from Geogebra

But I want to solve this problem using Synthetic or Trigonometry.

Someone solved this problem before when it was an equilateral triangle, but in this case it is difficult to solve because it is an isosceles triangle. Link: Conjecture about a point inside an equilateral triangle divided by integer angles

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This is the process of solving problems through drawing with @dfnu ideas.

Using regular pentagons and equilateral triangles, we can create the isosceles triangle and angles we are looking for. I think the proof will be complete when $\square ACGR $ shows an inscribed rectangle.

Answer 1

This question seems poorly received but I found it interesting. So I just provide a short solution.

By the trigonometric corollary of Ceva's theorem:

$$\frac{sin(54^\circ)}{sin(12^\circ)}\cdot\frac{sin(6^\circ)}{sin(42^\circ)}\cdot\frac{sin(x)}{sin(y)} = 1$$ And by the angle sum of a triangle, $$x+y = 66^\circ$$

$$\frac{sin(x)}{sin(66^\circ - x)} = \frac{sin(12^\circ)\cdot sin(42^\circ)}{sin(54^\circ)\cdot sin(6^\circ)}$$

Now this simplifies very easily if you know the compound angle identities and the trigonometric ratios for $18^\circ$ and $30^\circ$.

$$ \frac{2sin(6^\circ)cos(6^\circ)sin(42^\circ)}{sin(6^\circ)sin(54^\circ)} $$ $$ \frac{2sin(42^\circ)cos(6^\circ)}{cos(36^\circ)} $$

Do note that now I multiply the fraction by $\frac{sin(24^\circ)}{sin(24^\circ)}$ since it can be observed that helps in simplifying the fraction. Then I used the product to sum identities to get it in terms of the ratios in $18^\circ$ and $30^\circ$, which are more familiar.

$$ \frac{sin(42^\circ)[sin(30^\circ) + sin(18^\circ)]}{cos(36^\circ)sin(24^\circ)} $$

$$ \frac{sin(42^\circ)}{sin(24^\circ)} $$

$$ \frac{sin(x)}{sin(66^\circ - x)} = \frac{sin(42^\circ)}{sin(24^\circ)} $$ $$\fbox{x=42 degrees}$$

Answer 2

I think you are on the right track, now. To complete the proof, note first that $\triangle APQ$ is isosceles, with $\measuredangle PQA = \measuredangle PAQ = 6^\circ$. Therefore $\triangle BPQ \cong \triangle BCQ$ by ASA criterion. This in turn implies that $\square BPCQ$ is cyclic, with two opposite equal sides, hence an isosceles trapezoid. So $PC\parallel BQ$, from which you get your result, that is $\measuredangle BCP = \measuredangle CBQ = 42^\circ$.