Question

In the figure, not drawn to scale, ABCD is a trapezium. BF is a straight line. Find $\angle DCE$ and $\angle ADE$

• A. ${72}^{\circ },{152}^{\circ }$
• B. ${80}^{\circ },{152}^{\circ }$
• C. ${70}^{\circ },{142}^{\circ }$
• D. ${82}^{\circ },{150}^{\circ }$

Answer 1

The correct option is A ${72}^{\circ },{152}^{\circ }$

BF is a straight line.
So, $\angle BCF=180$
$\angle DCB+\angle DCE+\angle ECF=180$
$64+\angle DCE+44=180$
$\angle DCE=180-108$
$\angle DCE=72$
DC=DE
So, $\angle DCE=\angle DEC=72$
So, in \triangle CDE, $\angle DCE+\angle CDE+\angle DEC=180$
$\angle CDE+72+72=180$
$\angle CDE=36$
ABCD is a trapezium.and $AD\parallel BC$
So, $\angle BCD+\angle CDA=180$
$64+\angle CDA=180$
$\angle CDA=116$
So, $\angle ADE=\angle CDA+\angle DCE=116+36=152$
Answer (A) 72,152