Question

In triangle ABC, $\overline{AC}=\overline{CD}$ and $\measuredangle CAB-\measuredangle ABC={30}^0$. Then $\measuredangle BAD$ is:

• A. ${30}^0$
• B. ${20}^0$
• C. $22{\frac{1}{2}}^0$
• D. ${15}^0$

Answer 1

The correct option is D ${15}^0$

$\overline{AC}=\overline{CD}\therefore \angle CAD=\angle CDA$

$\angle CDA=\angle DAB+\angle DBA$ [$\because$ external angle = sum of two opposite internal angles of a triangle]

$\angle CDA=\angle DAB+\angle ABC$ [$\because \angle DBA=\angle ABC$]

$\angle CAD=\angle DAB+\angle ABC$

$\angle CAD+\angle DAB=2\angle BAD+\angle ABC\angle CAB-\angle ABC=2\angle BAD30°=2\angle BAD\therefore \angle BAD=15°$