Question

In the given figure, if $\angle AOC$ and $\angle BOC$ are in ratio 2 : 3, find $\angle AOC$, $\angle COB$, $\angle BOD$ and $\angle AOD$.

• A. ${72}^{\circ },{108}^{\circ },{72}^{\circ }and{108}^{\circ }$
• B. ${45}^{\circ },{100}^{\circ },{45}^{\circ }and{100}^{\circ }$
• C. ${30}^{\circ },{30}^{\circ },{120}^{\circ }and{120}^{\circ }$
• D. ${50}^{\circ },{120}^{\circ },{50}^{\circ }and{120}^{\circ }$

Answer 1

The correct option is A

${72}^{\circ },{108}^{\circ },{72}^{\circ }and{108}^{\circ }$

Given: $\angle AOC:\angle BOC=2:3$
Let measure of $\angle AOC$ and $\angle BOC$ be 2x and 3x.
Here, $\angle AOC$ and $\angle BOC$ form a linear pair of angles.
$\Rightarrow \angle AOC+\angle BOC={180}^{\circ }$
$\Rightarrow 2x+3x={180}^{\circ }$
$\Rightarrow 5x={180}^{\circ }$
$\Rightarrow x={36}^{\circ }$

$\therefore \angle AOC=2x=2\times {36}^{\circ }={72}^{\circ }$
$\angle BOC=3x=3\times {36}^{\circ }={108}^{\circ }$

We know that when two lines intersect, the vertically opposite angles are equal.
$\Rightarrow \angle BOD=\angle AOC={72}^{\circ }$
$\Rightarrow \angle AOD=\angle BOC={108}^{\circ }$