Question

In the given figure, AB = AC, BC = CD and $\angle$ A = ${40}^{\circ }$, find $\angle$ACD

• A. ${20}^{\circ }$
• B. ${25}^{\circ }$
• C. ${30}^{\circ }$
• D. ${35}^{\circ }$

Answer 1

The correct option is C ${30}^{\circ }$

$AB=AC$
$\therefore \angle ABC=\angle ACB$(Angles opposite to equal sides)
But $\angle ABC+\angle ACB+\angle BAC=180°$(Angles of a triangle)
$\Rightarrow \angle ABC+\angle ABC+{40}^{\circ }=180°$
$\Rightarrow 2\angle ABC={180}^{\circ }-{40}^{\circ }={140}^{\circ }$
$\therefore \angle ABC=\frac{{140}^{\circ }}{2}={70}^{\circ }$
In $△$ BDC, BC = CD
$\therefore \angle ABC=\angle BDC$ (Each $70°)$
In $△$ ADC,
Ext. $\angle BDC=\angle DAC+\angle DCA$
${40}^{\circ }+x={70}^{\circ }$
$x={30}^{\circ }$