Question

If in the following figure, AC = CD, AD = BD and $\angle C={58}^{\circ }$, then the measure of angle CAB is

• A. ${90}^{\circ }$
• B. ${91.5}^{\circ }$
• C. ${92.5}^{\circ }$
• D. ${93}^{\circ }$

Answer 1

The correct option is B

${91.5}^{\circ }$

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, $\angle ACD=50°$

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

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

${58}^{\circ }+2\angle CAD={180}^{\circ }$

$\angle CAD=\angle CDA={61}^{\circ }$

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)

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

$2\angle DAB={61}^{\circ }$

$\angle DAB={30.5}^{\circ }$

In triangle ABC,

$\angle CAB=\angle CAD+\angle DAB$

$\therefore \angle CAB={61}^{\circ }+{30.5}^{\circ }={91.5}^{\circ }$