Question

Find the angles marked with a question mark shown in the figure.

Answer 1

In parallelogram ABCD,

$CE\perp ABandCF\perp AD\angle BCE={40}^{\circ }$
In $\Delta BCE$,
$\angle BCE+\angle CEB+\angle EBC={180}^{\circ }$
(Sum of angles of a triangle)
$\Rightarrow {40}^{\circ }+{90}^{\circ }+\angle EBC={180}^{\circ }\Rightarrow {130}^{\circ }+\angle EBC={180}^{\circ }\Rightarrow \angle EBC={180}^{\circ }-{130}^{\circ }={50}^{\circ }or\angle B={50}^{\circ }$
But $\angle D=\angle B$ (Opposite angles)
$\therefore \angle D={50}^{\circ }OR\angle ADC={50}^{\circ }$
Similarly in $\Delta DCF$.
$\angle DCF+\angle CFD+\angle FDC={180}^{\circ }\Rightarrow DCF+{90}^{\circ }+{50}^{\circ }={180}^{\circ }\Rightarrow \angle DCF+{14}^{\circ }={180}^{\circ }\Rightarrow \angle DCF={180}^{\circ }-{140}^{\circ }={40}^{\circ }$
But $\angle C+\angle B={180}^{\circ }$
(Sum of adjacent angles)
$\Rightarrow \angle BCE+\angle ECF+\angle DCF+\angle B={180}^{\circ }\Rightarrow \angle ECF+{40}^{\circ }+{50}^{\circ }={180}^{\circ }\Rightarrow {40}^{\circ }+\angle ECF+{40}^{\circ }+{50}^{\circ }={180}^{\circ }\Rightarrow \angle ECF+{130}^{\circ }={180}^{\circ }\Rightarrow \angle ECF={180}^{\circ }-{130}^{\circ }={50}^{\circ }\therefore \angle ECF={50}^{\circ }$