Question

The sides BC, CA and AB of $\Delta ABC$ are produced in order to form exterior angles $\angle ACD$ , $\angle ABG$, and $\angle CAE$ then $\angle ACD+\angle CAE+\angle ABG$ is

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

Answer 1

The correct option is B ${360}^{\circ }$

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

$\Rightarrow$ $\angle ACD+\angle CAE+\angle ABG$
$=({180}^{\circ }-z)+({180}^{\circ }-x)+({180}^{\circ }-y)$

$\Rightarrow$ $\angle ACD+\angle CAE+\angle ABG$
$={540}^{\circ }-(x+y+z)={540}^{\circ }-{180}^{\circ }={360}^{\circ }$