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

  • Context: MHB 
  • Thread starter Thread starter maxkor
  • Start date Start date
  • Tags Tags
    Angle Triangle
Click For Summary

Discussion Overview

The discussion revolves around finding the angle $\angle PCA$ in triangle ABC, where point $N$ lies on segment $BC$ and point $P$ lies on segment $AN$. The problem includes given angles and relationships within the triangle, with participants exploring various geometric properties and relationships. The discussion encompasses mathematical reasoning and exploratory problem-solving.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Post 1 introduces the problem and the known angles: $\angle ANB = 90^\circ$, $\angle PBA = 20^\circ$, $\angle PBC = 40^\circ$, and $\angle PCB = 30^\circ$.
  • Post 2 suggests constructing a diagram to visualize the problem and asks for the angle $\angle NPB$.
  • Post 3 claims $\angle NPB = 50^\circ$ and seeks the next steps.
  • Post 4 questions what $\angle BPA$ must be after establishing $\angle NPB$.
  • Post 5 proposes $\angle BPA = 30^\circ$ and seeks further clarification.
  • Post 6 states that $\angle BPA + 50^\circ = 180^\circ$, leading to $\angle BPA = 130^\circ$.
  • Post 7 updates the diagram with new angles $\theta_5 = 50^\circ$ and $\theta_6 = 130^\circ$ and asks what else can be filled in.
  • Post 8 suggests additional angles: $\angle PAB = 30^\circ$, $\angle APC = 120^\circ$, and $\angle NCP = 120^\circ$.
  • Post 9 reiterates the angles and asks for the identification of remaining angles.
  • Post 10 expresses uncertainty about finding $\angle PAC$.
  • Post 11 introduces $\angle PAC = \beta$ and establishes a relationship with other angles, leading to $\alpha + \beta = 60^\circ$.
  • Post 12 discusses expressing the area of triangle ANC in multiple ways, leading to a system of equations.
  • Post 13 mentions circular reasoning with identities and suggests that a different approach may be needed.
  • Post 14 proposes that point $P$ is the orthocenter of triangle ABC, leading to the conclusion that $\alpha = 20^\circ$.

Areas of Agreement / Disagreement

Participants express various viewpoints and approaches to the problem, with no clear consensus on the final angle $\angle PCA$ or the best method to derive it. Multiple competing views and methods remain unresolved.

Contextual Notes

The discussion includes assumptions about the relationships between angles and the properties of triangle ABC, which may not be universally accepted or verified. The exploration of identities and equations introduces complexity that remains unresolved.

maxkor
Messages
79
Reaction score
0
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.
 
Mathematics news on Phys.org
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:

$$\theta_1=90^{\circ},\,\theta_2=20^{\circ},\,\theta_3=40^{\circ},\,\theta_4=30^{\circ}$$

Can you find $\angle\text{NPB}$?
 
$\angle NPB=50^o$ what next?
 
maxkor said:
$\angle NPB=50^o$ what next?

Then what must $\angle\text{BPA}$ be?
 
$30^o$ and?
 
maxkor said:
$30^o$ and?

We should have:

$$\angle\text{BPA}+50^{\circ}=180^{\circ}$$

Hence:

$$\angle\text{BPA}=130^{\circ}$$

Do you see any other angles you can fill in?
 
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:

$$\theta_5=50^{\circ},\,\theta_6=130^{\circ}$$

What else can we fill in?
 
$\angle PAB=30^o,\angle APC=120^o, \angle NCP =120^o$ and?
 
maxkor said:
$\angle PAB=30^o,\angle APC=120^o, \angle NCP =120^o$ and?

What I find is:

$$\angle\text{PAB}=30^{\circ},\,\angle\text{NPC}=60^{\circ},\,\angle\text{APC}=120^{\circ}$$

We have two angles left to find...can you identify them?
 
  • #10
For example, I don't know how to find $\angle PAC$
 
  • #11
maxkor said:
For example, I don't know how to find $\angle PAC$

Let's let:

$$\angle\text{PAC}=\beta$$

So, we may state:

$$\alpha+\beta+120^{\circ}=180^{\circ}$$

or:

$$\alpha+\beta=60^{\circ}$$

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

$$x=\overline{NC}$$

$$y=\overline{AN}$$

$$z=\overline{AC}$$

By Pythagoras, we have:

$$x^2+y^2=z^2$$

Can you express the area of the right triangle in 3 different ways, so that we have 5 equations in 5 unknowns?
 
  • #12
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
 
  • #13
maxkor said:
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
 
  • #14
Using values as given by Mark's diagram, notice that $P$ is the orthocenter of $\triangle{ABC}$, so $\alpha=90^\circ-\theta_3-\theta_4=20^\circ$.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
14
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K