Question

In an isosceles trapezium $PQRS,\angle P$ and $\angle S$ are in the ratio $1:2$. Find the measure of all the angles.

Answer 1

REF Image

In an isosceles trapezium base angles are equal; $\angle P=\angle Q$.
Let $\angle P=x^{\circ }$ and $\angle S=2x^{\circ }$.

Since $\angle P+\angle S={180}^{\circ }$ (one pair of adjacent angles of a trapezium is supplementary), we get
$x^{\circ }+2x^{\circ }={180}^{\circ }$.
Thus, $3x^{\circ }={180}^{\circ }$ or $x^{\circ }={180}^{\circ }/3={60}^{\circ }$.

Hence $\angle P={60}^{\circ }$ and $\angle S=2\times {60}^{\circ }={120}^{\circ }$. Since $\angle P=\angle Q$, we get $\angle Q={60}^{\circ }$. But, $\angle Q+\angle R={180}^{\circ }$ (one pair of adjacent angles are supplementary).

Hence we also get $\angle R={180}^{\circ }-{60}^{\circ }={120}^{\circ }$.