Question

Two straight lines AB and CD intersect one another at the point O. If $\angle AOC+\angle COB+\angle BOD={274}^0$, then $\angle AOD$

• A. ${86}^0$
• B. ${90}^0$
  
• C. ${94}^0$
• D. ${137}^0$

Answer 1

The correct option is A

${86}^0$

Sum of angles at a point O= ${360}^0$
Sum of three angles $\angle AOC+\angle COB+\angle BOD={274}^0$
$\therefore$ Fourth angle $\angle AOD={360}^0-{274}^0={86}^0$