Question

In the figure given above, $AB$ is parallel to $CD$. If $\angle DCE=x$ and $\angle ABE=y$, then what is $\angle CEB$ equal to?

• A. $y-x$
• B. $\frac{(x+y)}{2}$
• C. $x+y-\left(\pi /2\right)$
• D. $x+y-\pi$

Answer 1

The correct option is D $x+y-\pi$

We draw $l\parallel AB$ through $E$.
We have $\angle 1=180-y^{\circ }$

and $\angle 2={180}^{\circ }-x^{\circ }$
We know $\angle 1+\angle 2+\angle BEC=180$
$\Rightarrow 180-x+180-y+\angle BEC=180$
$\Rightarrow \angle BEC=180-360+x+y$
$=(x+y-{180}^{\circ })$ or $(x+y-\pi )$