Question

Observe the figure given.

Compute the sume 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,
$\angle ABE=(150-x{)}^{\circ }and\angle CBD=(70-x{)}^{\circ }$

Taking the straight line ABC,
$(150-x)+(70-x)+x={180}^{\circ }$
$\Rightarrow x={220}^{\circ }-{180}^{\circ }={40}^{\circ }$

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

In $\Delta BCD,\angle BDC=x$ (Alternate angles)
$(70-x)+x+z={180}^{\circ }\Rightarrow z={110}^{\circ }$

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