Question

In the above figure, $\text{A}D$ = $\text{D}C$= $\text{C}B$ and the measure of $\angle DAC$ = ${40}^o$ . What is the measure of $\angle DCB$?

• A. ${80}^o$
• B. ${40}^o$
• C. ${20}^o$
• D. ${300}^o$

Answer 1

The correct option is C ${20}^o$

$\because AD=DC$

$⟹$ $\angle DAC=\angle ACD$ (By isosceles triangle property, angles opposite equal sides are equal).

In $\Delta ACD:$

${40}^{\circ }+{40}^{\circ }+\angle CDA={180}^{\circ }$ (By angle sum property of a triangle)

$⟹{80}^{\circ }+\angle CDA={180}^{\circ }$

$⟹\angle CDA={180}^{\circ }-{80}^o$

$⟹\angle CDA={100}^{\circ }$.

$\therefore \angle CDB={180}^{\circ }-{100}^{\circ }={80}^{\circ }$ $(\because$ linear pair, i.e. sum of angles on a straight line $={180}^{\circ })$.

In $\Delta DCB$ :

$\angle BDC=\angle DBC$ (By isosceles triangle property, angles opposite equal sides are equal).

$⟹\angle DBC={80}^{\circ }$.

$In\Delta DBC$

$⟹$${80}^{\circ }+{80}^{\circ }+\angle DCB={180}^{\circ }$ (By angle sum property of a triangle)

$⟹$$\angle DCB={180}^{\circ }-{160}^{\circ }$$={20}^{\circ }$.

Therefore, option $C$ is correct.