Finding $\angle APB$ in $PQR$ Triangle with $QA:AB:BR = 3:5:4$

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Discussion Overview

The discussion revolves around finding the angle $\angle APB$ in triangle $PQR$, where $\angle P=90^{\circ}$ and the segments on $QR$ are divided in the ratio $QA:AB:BR=3:5:4$. The scope includes mathematical reasoning and application of the cosine rule.

Discussion Character

  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • Participants describe triangle $PQR$ with specific properties, including $\angle P=90^{\circ}$ and $PQ=PR$.
  • One participant proposes specific lengths for segments $QA$, $AB$, and $BQ$ based on the given ratio, suggesting $QA=3$, $AB=5$, and $BQ=4$.
  • Using the cosine rule, calculations are presented for triangles $PQA$, $PRB$, and $APB$, leading to the conclusion that $\alpha = 45^\circ$.
  • Another participant reiterates the calculations and confirms the result, expressing appreciation for the contributions of others.
  • A later post hints at an alternative geometric approach to the problem without providing details.

Areas of Agreement / Disagreement

There is a general agreement on the calculations leading to $\angle APB = 45^\circ$, but the discussion includes a suggestion of an alternative approach that remains unexplored.

Contextual Notes

Participants rely on specific segment lengths derived from the ratio, and the calculations depend on the application of the cosine rule. The discussion does not resolve the potential for alternative methods to find $\angle APB$.

anemone
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Let $PQR$ be a triangle with $\angle P=90^{\circ}$ and $PQ=PR$. Let $A$ and $B$ be points on the segment $QR$ such that $QA:AB:BR=3:5:4$. Find $\angle APB$.
 
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anemone said:
Let $PQR$ be a triangle with $\angle P=90^{\circ}$ and $PQ=PR$. Let $A$ and $B$ be points on the segment $QR$ such that $QA:AB:BR=3:5:4$. Find $\angle APB$.
[sp]
If $QA=3$, $AB=5$ and $BQ=4$ then the other two sides of the triangle $PQR$ will both be $6\sqrt2$ units.Cosine rule in triangle $PQA$: $AP^2 = 9 + 72 - 2\cdot 3\cdot 6\sqrt2\cdot\dfrac1{\sqrt2} = 45$.

Cosine rule in triangle $PRB$: $BP^2 = 16 + 72 - 2\cdot 4\cdot 6\sqrt2\cdot\dfrac1{\sqrt2} = 40$.

Cosine rule in triangle $APB$: $\cos\alpha = \dfrac{40 + 45 - 25}{2\cdot\sqrt{40}\cdot\sqrt{45}} = \dfrac1{\sqrt2}$.

Therefore $\alpha = 45^\circ$.[/sp]
 

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Opalg said:
[sp]
If $QA=3$, $AB=5$ and $BQ=4$ then the other two sides of the triangle $PQR$ will both be $6\sqrt2$ units.Cosine rule in triangle $PQA$: $AP^2 = 9 + 72 - 2\cdot 3\cdot 6\sqrt2\cdot\dfrac1{\sqrt2} = 45$.

Cosine rule in triangle $PRB$: $BP^2 = 16 + 72 - 2\cdot 4\cdot 6\sqrt2\cdot\dfrac1{\sqrt2} = 40$.

Cosine rule in triangle $APB$: $\cos\alpha = \dfrac{40 + 45 - 25}{2\cdot\sqrt{40}\cdot\sqrt{45}} = \dfrac1{\sqrt2}$.

Therefore $\alpha = 45^\circ$.[/sp]
Well done Opalg (Yes) and thanks for participating!

Here is another approach that tackles the problem using the geometry route that I want to share with you:
View attachment 3122

Here is another approach that tackles the problem using the geometry route that I want to share with you:

If we rotate the triangle $PQR$ about $P$ by $90^{\circ}$, the point $Q$ goes to $R$.

Let $A'$ represent the image of point $A$ under this rotation. Then we have

$RA'=QA$ and $\angle PRA'=\angle PQR=45^{\circ}$ so $BRA'$ is a right-angled triangle with $RB:RA'=4:3$, $\therefore A'B=BA$.

It follows that $PABA"$ is a kite with $PA'=PA$ and $AB=BA'$. Therefore $PB$ is the angle bisector of $\angle APA'$. This implies $\angle APB=\dfrac{\angle APA'}{2}=45^{\circ}$.
 

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anemone said:
Let $PQR$ be a triangle with $\angle P=90^{\circ}$ and $PQ=PR$. Let $A$ and $B$ be points on the segment $QR$ such that $QA:AB:BR=3:5:4$. Find $\angle APB$.
let point Q be the origin so we may construct the following :
Q(0,0),A(3,0),B(8,0),R(12,0) and P(6,6)
slope of PA=2
slope of PB=-3
and we have :tan ($\angle $ APB)=1
that is $\angle $APB=$45^o$
 
Last edited:
Albert said:
let point Q be the origin so we may construct the following :
Q(0,0),A(3,0),B(8,0),R(12,0) and P(6,6)
slop of PA=2
slop of PB=-3
and we have :tan ($\angle $ APB)=1
that is $\angle $APB=$45^o$

Good job, Albert! And thanks for participating!:)
 

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