Question

In fig 10.49, $\text{ABCD}$ is parallelogram. $\text{CE}\perp \text{AB,CF}\perp \text{AD}\text{and}\angle \text{BCE = 4}{\text{0}}^{\text{o}}\text{.}$ Find the values of x, y and z.

Answer 1

$\angle CEB={90}^{\circ }$, In $△BCE,40+90+x={180}^{\circ }$

$\Rightarrow {130}^{\circ }+x={180}^{\circ }$

$\Rightarrow x={50}^{\circ }$

But ABCD $\angle B=\angle D={50}^{\circ }$

In $△DFC$, $Z+90+\angle DCF={180}^{\circ }$

$\Rightarrow 50+90+\angle DCF={180}^{\circ }$

$\Rightarrow \angle DCF={40}^{\circ }$

But $\angle B\pm \angle C={180}^{\circ }=x+\angle DCF+y+{40}^{\circ }={180}^{\circ }$

$=50+40+y+{40}^{\circ }={180}^{\circ }$

$=y+{130}^{\circ }={180}^{\circ }$

$=y={50}^{\circ }$