Question

In the figure, $\angle BAD=3\angle DBA$, find $\angle ABC$.

• A. ${58}^0$
• B. ${59}^0$
• C. ${96}^0$
• D. ${86}^0$

Answer 1

Answer: (A)

The correct option is A ${58}^0$

In $△ABD$, let $\angle DBA=\theta$

$3\theta +\theta +104=180$

$\Rightarrow 4\theta ={76}^{\circ }$

$\Rightarrow \theta ={19}^{\circ }$

${104}^{\circ }+\angle BDC={180}^{\circ }$(sum of angles on a straight line=${180}^{\circ })$

$\Rightarrow \angle BDC={76}^{\circ }$

In $△BDC:$

$\Rightarrow {76}^{\circ }+{65}^{\circ }+\angle DBC={180}^{\circ }$

$\Rightarrow \angle DBC={39}^{\circ }$

$\angle ABC=\angle ABD+\angle DBC$

$=\theta +{39}^{\circ }$

$={19}^{\circ }+{39}^{\circ }$

$={58}^{\circ }$