Question
Find the remaining angles in the following trapeziums -
✓
Answer
Since $A B\ \|\ D C$, and $A D$ is a tranversal, then
$\angle A +\angle D =180^{\circ}$ $\qquad$ (Sum of angles on the same side of the transversal)
$135^{\circ}+\angle D =180^{\circ}$
$\angle D =180^{\circ}-135^{\circ}$
$\angle D =45^{\circ}$
Also, since $AB\ \|\ DC$, and BC is a tranversal, then
$\angle B +\angle C =180^{\circ}$ $\qquad$ (Sum of angles on the same side of the transversal)
$\begin{array}{l}105^{\circ}+\angle C=180^{\circ} \\\angle C=180^{\circ}-105^{\circ} \\\angle C=75^{\circ}\end{array}$
Since $PQ \| SR$, and PS is a tranversal, then
$\angle P +\angle S =180^{\circ}$ $\qquad$ (Sum of angles on the same side of the transversal)
$\angle P+100^{\circ}=180^{\circ}$
$\angle P =180^{\circ}-100^{\circ}=80^{\circ}$.
$\angle S =\angle R =100^{\circ}$ $\qquad$ (Angles opposite to equal sides are equal)
Also, since $PQ \| SR$, and QR is a tranversal, then
$\angle Q +\angle R =180^{\circ}$ $\qquad$ (Sum of angles on the same side of the transversal)
$\angle Q+100^{\circ}=180^{\circ}$
$\angle Q=180^{\circ}-100^{\circ}=80^{\circ}$.
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