Question

From the given figure :

Find the sum of x, y and z.

• A. ${180}^{\circ }$
• B. ${70}^{\circ }$
• C. ${80}^{\circ }$
• D. ${190}^{\circ }$

Answer 1

The correct option is D

${190}^{\circ }$

From the figure BE is parallel to DC

$\angle EBC+\angle z=(180{)}^{\circ }$ (Co interior angles)

${70}^{\circ }+\angle z=(180{)}^{\circ }$

$\angle z=(180-70{)}^{\circ }={110}^{\circ }$

$\angle ABE=\angle z={110}^{\circ }$ (corresponding angles)

$\angle x=\angle ABD-\angle ABE$

$={150}^{\circ }-{110}^{\circ }={40}^{\circ }$

Since AE || BD, y = x as they are alternate angles.

$\angle y={40}^{\circ }$

$\therefore$ The required sum $=x+y+z={40}^{\circ }+{40}^{\circ }+{110}^{\circ }={190}^{\circ }$