Question

In the figure, lines $AB$ and $CD$ intersect at $O$. If $\angle AOD:\angle DOB=4:5,$ then $\angle COB$ is :

• A. ${80}^{\circ }$
• B. ${100}^{\circ }$
• C. ${90}^{\circ }$
• D. ${70}^{\circ }$

Answer 1

The correct option is A ${80}^{\circ }$

Let $\angle AOD=4x$ and $\angle DOB=5x$

$\angle AOD+\angle DOB={180}^{\circ }$ (Adjacent angle on straight line are supplementary)

$\Rightarrow 4x+5x={180}^{\circ }\Rightarrow 9x={180}^{\circ }\Rightarrow x=\frac{{180}^{\circ }}{9}={20}^{\circ }$

$\angle AOD=4x$

$\Rightarrow \angle AOD=4\times {20}^{\circ }={80}^{\circ }$

Now $\angle COB=\angle AOD$ (vertically opposite angles)

$\Rightarrow \angle COB={80}^{\circ }$