In the diagram, XYZ is a straight line. AY = 20 m, AZ = 12 m and YZ = 16 m. The ratio of XY : YZ is 1 : 4.

1. Identify the longest side and compare the squares of the side lengths

In triangle $AYZ$,

$AY=20$, $AZ=12$ and $YZ=16$.

The longest side is $AY$.

$AZ^2+YZ^2=12^2+16^2=144+256=400$

$AY^2=20^2=400$

2. Conclude using the converse of Pythagoras' theorem

Since

$AZ^2+YZ^2=AY^2,$

by the converse of Pythagoras' theorem, triangle $AYZ$ is right-angled.

The right angle is between $AZ$ and $YZ$, so it is at $Z$.

1. Relate $\angle AYX$ to the known right-angled triangle $AYZ$ and find the cosine value

Since $X,Y,Z$ are on a straight line, $YX$ is the extension of $YZ$.

In right-angled triangle $AYZ$,
$\cos\angle AYZ=\frac{YZ}{AY}=\frac{16}{20}=\frac{4}{5}.$

Hence
$\cos\angle AYX=-\frac{4}{5}.$

1. Use the area formula for triangle $AYW$ to find $\sin \angle AYW$

Since $YW=YX$ and $XY:YZ=1:4$ with $YZ=16$, we have
$YX=\frac14(16)=4.$
For triangle $AYW$,
$\text{Area}=\frac12(AY)(YW)\sin \angle AYW.$
Given the area is $24\text{ m}^2$,
$24=\frac12(20)(4)\sin \angle AYW$
$24=40\sin \angle AYW$
$\sin \angle AYW=\frac{24}{40}=\frac35.$

2. Show that an acute angle can give the same area

If
$\sin \angle AYW=\frac35,$
then one possible value is
$\angle AYW=\sin^{-1}\left(\frac35\right)=36.9^\circ.$
This angle is acute. Therefore a triangle $AYW$ with $YW=YX$ and the same area can have angle $Y$ acute.