The largest angle of the quadrilateral PQRS is 120°
Step-by-step explanation:
Given that
[tex]\frac{\angle P}{2} =\frac{\angle Q}{3}=\frac{\angle R}{4}[/tex]
3P=2Q (1)
4Q=3R
Q=3R/4
Substitute the value of <Q in (1)
[tex]3P=2(\frac{3R}{4} )=\frac{3R}{2} \\6P=3R\ -->(2)[/tex]
<P=3R/6=R/2
Comparing (1) and (2)
we get 6P=4Q=3R
In a cyclic quadrilateral, opposite angles are supplementary i.e their sum is 180 degree
<P and <R are opposite angles.
Hence [tex]<P+<R=180\°[/tex]
<R/2+<R=180
[tex]\frac{3R}{2} =180\\R=120\°\\<P=R/2=120/2=60\°[/tex]
[tex]<Q=\frac{3R}{4} =\frac{3\times 120}{4} =90\°[/tex]
<S and <Q are opposite angles. Hence
<S=180-<Q=180-90=90°
<P=60°
<Q=90°
<R=120°
<S=90°
The largest angle of the quadrilateral is 120°(<R)
