Question

In the figure if $\angle$ $BDC=30$ ${}^{\circ }$, $\angle$ $CBA=110$ ${}^{\circ }$, then find $\angle BCA$:

• A. ${20}^{\circ }$
• B. ${40}^{\circ }$
• C. ${35}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is B ${40}^{\circ }$

$\angle BAC=\angle BDC$ ($\angle s$ in the same segment)$=30$ ${}^{\circ }$ (given)
$\angle BAC+\angle CBA+\angle BCA={180}^{\circ }$ ($\angle$ s of $\Delta ABC$)
$\Rightarrow {30}^{\circ }+{110}^{\circ }+\angle BCA={180}^{\circ }$
$\Rightarrow \angle BCA={180}^{\circ }-{30}^{\circ }-{110}^{\circ }={40}^{\circ }$