Question

If ABCD is a trapezium, find $\angle A$.

• A. ${135}^{\circ }$
• B. ${125}^{\circ }$
• C. ${115}^{\circ }$
• D. ${35}^{\circ }$

Answer 1

The correct option is A ${135}^{\circ }$

The angles on the non parallel sides are supplementary.
$\angle A+\angle B={180}^{\circ }$
$3x+{30}^{\circ }+x+{10}^{\circ }={180}^{\circ }$
$4x+{40}^{\circ }={180}^{\circ }$
$4x={180}^{\circ }-{40}^{\circ }$
$4x={140}^{\circ }$
$x={35}^{\circ }$
$\therefore \angle A=3x+{30}^{\circ }$
$\angle A=3\times {35}^{\circ }+{30}^{\circ }$
$\angle A={135}^{\circ }$