Question

Find all the angles in the given quadrilateral below

Answer 1

We know that $\angle P+\angle Q+\angle R+\angle S={360}^{\circ }$ (angle sum property). Hence
$x+2x+3+x+3x-7={360}^{\circ }$.
This gives $7x-4={360}^{\circ }$ or $7x={364}^{\circ }$.
Therefore $x=364/7={52}^{\circ }$.
We obtain, $\angle P={52}^{\circ };angleR={52}^{\circ };\angle Q=2x+3=2\times 52+3=(104+3)={107}^{\circ };\angle S=3\times {52}^{\circ }-7^{\circ }={156}^{\circ }-7^{\circ }={149}^{\circ }$. Check that $\angle P+\angle Q+\angle R+\angle S={360}^{\circ }$.