Finding $\angle PCA$ in Triangle ABC with $\angle ANB = 90^\circ$

[maxkor]

In triangle ABC point $N \in BC$ , point $P \in AN$. Let
$\angle ANB = 90$°
$\angle PBA = 20$°
$\angle PBC = 40$°
$\angle PCB = 30$°.
Find $\angle PCA$.
I don't know how to solve that. Maby it was.

 

[MarkFL]

Let's begin by constructing a diagram:

\begin{tikzpicture}[blue]
\coordinate (A) at (0,0);
\coordinate (B) at (2,5);
\coordinate (C) at (8,0);
\coordinate (N) at (3.28,3.93);
\coordinate (P) at (2,2.4);
\draw[blue, ultra thick] (A) -- (B) -- (C) -- cycle;
\draw[blue, ultra thick] (A) -- (N);
\draw[blue, ultra thick] (B) -- (P);
\draw[blue, ultra thick] (P) -- (C);
\path (A) node[below left] {A} -- (B) node[above] {B} -- (C) node[below right] {C} -- (N) node[above right] {N} -- (P) node[below=5pt] {P} -- (C) node[above=5pt, left=25pt] {$\alpha$} -- (N) node[left=5pt] {$\theta_1$} -- (B) node[below=30pt,left=-2pt] {$\theta_2$} -- (B) node[below=12pt,right=-2pt] {$\theta_3$} -- (C) node[above=20pt, left=25pt] {$\theta_4$};
\end{tikzpicture}

where:

\(\displaystyle \theta_1=90^{\circ},\,\theta_2=20^{\circ},\,\theta_3=40^{\circ},\,\theta_4=30^{\circ}\)

Can you find $\angle\text{NPB}$?

 

[maxkor]

$\angle NPB=50^o$ what next?

 

[MarkFL]

$\angle NPB=50^o$ what next?

Then what must $\angle\text{BPA}$ be?

 

[maxkor]

$30^o$ and?

 

[MarkFL]

$30^o$ and?

We should have:

\(\displaystyle \angle\text{BPA}+50^{\circ}=180^{\circ}\)

Hence:

\(\displaystyle \angle\text{BPA}=130^{\circ}\)

Do you see any other angles you can fill in?

 

[MarkFL]

Okay, we now have:

\begin{tikzpicture}[blue]
\coordinate (A) at (0,0);
\coordinate (B) at (2,5);
\coordinate (C) at (8,0);
\coordinate (N) at (3.28,3.93);
\coordinate (P) at (2,2.4);
\draw[blue, ultra thick] (A) -- (B) -- (C) -- cycle;
\draw[blue, ultra thick] (A) -- (N);
\draw[blue, ultra thick] (B) -- (P);
\draw[blue, ultra thick] (P) -- (C);
\path (A) node[below left] {A} -- (B) node[above] {B} -- (C) node[below right] {C} -- (N) node[above right] {N} -- (P) node[below=5pt] {P} -- (C) node[above=5pt, left=25pt] {$\alpha$} -- (N) node[left=5pt] {$\theta_1$} -- (B) node[below=30pt,left=-2pt] {$\theta_2$} -- (B) node[below=12pt,right=-2pt] {$\theta_3$} -- (C) node[above=20pt, left=25pt] {$\theta_4$} -- (P) node[above=22pt, right=0pt] {$\theta_5$} -- (P) node[above=5pt, left=0pt] {$\theta_6$};
\end{tikzpicture}

where:

\(\displaystyle \theta_5=50^{\circ},\,\theta_6=130^{\circ}\)

What else can we fill in?

 

[maxkor]

$\angle PAB=30^o,\angle APC=120^o, \angle NCP =120^o$ and?

 

[MarkFL]

$\angle PAB=30^o,\angle APC=120^o, \angle NCP =120^o$ and?

What I find is:

\(\displaystyle \angle\text{PAB}=30^{\circ},\,\angle\text{NPC}=60^{\circ},\,\angle\text{APC}=120^{\circ}\)

We have two angles left to find...can you identify them?

 

[maxkor]

For example, I don't know how to find $\angle PAC$

 

[MarkFL]

For example, I don't know how to find $\angle PAC$

Let's let:

\(\displaystyle \angle\text{PAC}=\beta\)

So, we may state:

\(\displaystyle \alpha+\beta+120^{\circ}=180^{\circ}\)

or:

\(\displaystyle \alpha+\beta=60^{\circ}\)

Now, consider the right triangle $\triangle\text{ANC}$. Let:

\(\displaystyle x=\overline{NC}\)

\(\displaystyle y=\overline{AN}\)

\(\displaystyle z=\overline{AC}\)

By Pythagoras, we have:

\(\displaystyle x^2+y^2=z^2\)

Can you express the area of the right triangle in 3 different ways, so that we have 5 equations in 5 unknowns?

 

[maxkor]

Yes but when I use
$S=0.5xy$
$S=0.5xz\sin(60-\alpha)$
$S=0.5zysin(\alpha+30)$

$sin^2(60-\alpha)+sin^2(\alpha+30)=1$ it is identity

 

[MarkFL]

Yes but when I use
$S=0.5xy$
$S=0.5xz\sin(60-\alpha)$
$S=0.5zysin(\alpha+30)$

$sin^2(60-\alpha)+sin^2(\alpha+30)=1$ it is identity

Yes, I keep going in circles with identities...I suppose another approach is needed. :D

 

[Greg]

Using values as given by Mark's diagram, notice that $P$ is the orthocenter of $\triangle{ABC}$, so $\alpha=90^\circ-\theta_3-\theta_4=20^\circ$.