Question

The sides BC, CA and AB of $\Delta ABC$ are produced in order to form exterior angles $\angle ACD$ , $\angle EAB$, and $\angle CBF$. If $x+y+z={180}^{\circ }$ then $\angle ACD+\angle CBF+\angle EAB$ is

• A. ${180}^{\circ }$
• B. ${270}^{\circ }$
• C. ${360}^{\circ }$
• D. ${540}^{\circ }$

Answer 1

The correct option is C

${360}^{\circ }$

Given $x+y+z={180}^{\circ }$

$\angle ACD+\angle CBF+\angle EAB$
$=({180}^{\circ }-z)+({180}^{\circ }-x)+({180}^{\circ }-y)$
$={540}^{\circ }-(x+y+z)$
$={540}^{\circ }-{180}^{\circ }$
$={360}^{\circ }$