Question

$ABandBC$ are the sides of a regular pentagon. $CDandDE$ are the sides of a regular hexagon. If $OBandOD$ are the angle bisectors of $\angle ABCand\angle CDE,$ $and\angle BOD={126}^{\circ },$ then find the measure of $\angle BCD$.

• A. ${60}^{\circ }$
• B. ${120}^{\circ }$
• C. ${100}^{\circ }$
• D. ${80}^{\circ }$

Answer 1

The correct option is B ${120}^{\circ }$


$\because ABandBC$ are the sides of regular pentagon

$\angle ABC=\frac{(5-2)\times {180}^{\circ }}{5}={108}^{\circ }$

$\Rightarrow$ Each angle of a regular polygon
$=\frac{(n-2)\times {180}^{\circ }}{n}$

$\therefore \angle OBC=\frac{\angle ABC}{2}={54}^{\circ }$

Similarly,
$\angle BCD=\frac{(6-2)\times {180}^{\circ }}{6}={120}^{\circ }$

$\therefore \angle ODC=\frac{\angle BCD}{2}={60}^{\circ }$

Now,
$\angle OBC+\angle ODC+\angle BOD+\angle BCD={360}^{\circ }$
[Sum of all the angles of a quadrilateral $={360}^{\circ }$]

${54}^{\circ }+{60}^{\circ }+{126}^{\circ }+\angle BCD={360}^{\circ }$

$⟹\angle BCD={360}^{\circ }-({54}^{\circ }+{60}^{\circ }+{126}^{\circ })$

$⟹\angle BCD={120}^{\circ }$