Question

In the given figure, side $BA$ of $△ABC$ has been produced to $D$ such that $CD=CA$ and side $CB$ has been produced to $E$. If $\angle BAC={106}^0$ and $\angle ABE={128}^0$, find $\angle BCD$

• A. ${54}^0$
• B. ${52}^0$
• C. ${74}^0$
• D. ${60}^0$

Answer 1

The correct option is A ${54}^0$

In $△ADC$
$\angle CAB+\angle CAD=180$ (Linear pair)
$106+\angle CAD=180$
$\angle CAD={74}^{\circ }$
Given, $AC=CD$
$\angle CAD=\angle CDA={74}^{\circ }$ (Isosceles triangle property)
Also, $\angle ABC+\angle ABE=180$ (Linear pair)
$\angle ABC+128=180$
$\angle ABC=52$
Now, in $△BDC$
$\angle DBC+\angle DCB+\angle BDC=180$
$52+74+\angle DCB=180$
$\angle DCB={54}^{\circ }$