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

[anemone]

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 $\angle APB$.

 

[Opalg]

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]

View attachment 3121​

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]

 

[anemone]

[sp]

View attachment 3121​

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:

Spoiler

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 $\angle APB=\dfrac{\angle APA'}{2}=45^{\circ}$.

 

[Albert1]

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 $\angle APB$.

Spoiler

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=$45^o$

 

[anemone]

Spoiler

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!:)