Question

If in the following figure, AC = CD, AD = BD and $\angle C={52}^{\circ }$, then the measure of angle DAB is

• A. ${30}^{\circ }$
• B. ${32}^{\circ }$
• C. ${34}^{\circ }$
• D. ${36}^{\circ }$

Answer 1

The correct option is B

${32}^{\circ }$

In $△$ACD, AC = CD (given)

$⟹\angle CAD=\angle CDA$ (In a triangle, if two sides are equal, then the angles opposite to these sides are also equal)

Given, $\angle C=\angle ACD={52}^{\circ }$

As the sum of all angles of a triangle is ${180}^{\circ }$, considering triangle ACD, we get,

$\angle ACD+\angle CDA+\angle CAD={180}^{\circ }$

${52}^{\circ }+2\angle CAD={180}^{\circ }$

$\angle CAD+\angle CDA={64}^{\circ }$

Now, we must have,

$\angle CDA=\angle DAB+\angle DBA$ (Since exterior angle is the sum of opposite interior angles)

But $\angle DAB=\angle DBA$ (Since AD = DB)

$\therefore \angle DAB+\angle DAB=\angle CDA$

$2\angle DAB={64}^{\circ }$

$\angle DAB={32}^{\circ }$