Question

In the figure, $\angle BAD=4\angle DBA$, what is the measure of $\angle ABC$ and x in degrees?

• A. ${15}^{\circ },{45}^{\circ }$
• B. ${60}^{\circ },{45}^{\circ }$
• C. ${30}^{\circ },{60}^{\circ }$
• D. ${60}^{\circ },{30}^{\circ }$

Answer 1

The correct option is B ${60}^{\circ },{45}^{\circ }$

$Given\angle BAD=4\angle DBA\angle ADB={105}^{\circ },\angle BCD={60}^{\circ }$

In $\Delta ADB$,
$\angle DBA+\angle ADB+\angle BAD={180}^{\circ }\left[Anglesumproperty\right]\Rightarrow \angle DBA+{105}^{\circ }+4\angle DBA={180}^{\circ }\Rightarrow {105}^{\circ }+5\angle DBA={180}^{\circ }\Rightarrow 5\angle DBA={180}^{\circ }-{105}^{\circ }\Rightarrow \angle DBA=\frac{{75}^{\circ }}{5}={15}^{\circ }In\Delta DBC,\angle DBC+\angle BCD+\angle CDB={180}^{\circ }\left[Anglesumproperty\right][\because \angle CDB+\angle BDA={180}^{\circ }So,\angle CDB={180}^{\circ }-{105}^{\circ }={75}^{\circ }]\Rightarrow x+{60}^{\circ }+{75}^{\circ }={180}^{\circ }\Rightarrow x={180}^{\circ }-{135}^{\circ }={45}^{\circ }In\Delta ABC,\Rightarrow \angle ABC=\angle DBA+\angle DBC={15}^{\circ }+x^{\circ }={15}^{\circ }+{45}^{\circ }\angle B={60}^{\circ }$