Question

In the given figure, AB and CD are two straight lines intersecting at O. If $\angle \text{AOC and}\angle BOC\text{are 1:3 in ratio, then find}$
$\angle \text{BOD}and\angle \text{AOD, respectively}.$

• A. ${135}^{\circ }\text{and}{135}^{\circ }$
• B. ${45}^{\circ }\text{and}{45}^{\circ }$
• C. ${135}^{\circ }\text{and}{45}^{\circ }$
• D. ${45}^{\circ }\text{and}{135}^{\circ }$

Answer 1

The correct option is D ${45}^{\circ }\text{and}{135}^{\circ }$


$\angle AOCand\angle BOC\text{are supplementary angles.}$

$\therefore \angle AOC+\angle BOC={180}^{\circ }$

$Now,\angle AOC:\angle BOC=1:3$

$\text{Let}\angle \text{AOC = 1x and}\angle BOC=3x$

So, we can write:
$\angle AOC+\angle BOC={180}^{\circ }$

$\Rightarrow 1x+3x={180}^{\circ }$

$\Rightarrow 4x={180}^{\circ }$

$\Rightarrow x={45}^{\circ }$

$Thus,\angle AOC=x={45}^{\circ }$
$And\angle BOC=3x=3\times {45}^{\circ }={135}^{\circ }$

$\text{Now,}\angle \text{BOD}and\angle AOC\text{are}$
$\text{vertical angles.}$

$\therefore \angle BOD=\angle AOC={45}^{\circ }$

$\text{Now,}\angle \text{AOD}and\angle BOC\text{are}$
$\text{vertical angles.}$

$\therefore \angle AOD=\angle BOC={135}^{\circ }$