Question

Question 93

In the given figure, OB is perpendicular to OA and $\angle$BOC = ${49}^{\circ }$. Find $\angle$ AOD

• A. ${139}^{\circ }$
• B. ${49}^{\circ }$
• C. ${311}^{\circ }$
• D. ${149}^{\circ }$

Answer 1

The correct option is A

${139}^{\circ }$

From the given figure, we have
$\angle$DOB + $\angle$BOC = ${180}^{\circ }$ __[ (linear-pair)
or $\angle$DOB + ${49}^{\circ }$ = ${180}^{\circ }$
or, $\angle$DOB = ${180}^{\circ }$ - ${49}^{\circ }$ = ${131}^{\circ }$
Now, $\angle$DOB + $\angle$BOA + $\angle$AOB = ${360}^{\circ }$ ____ [∴Sum of all angles around a point is ${360}^{\circ }$ ]
or, ${131}^{\circ }$ + ${90}^{\circ }$ + $\angle$AOD = ${360}^{\circ }$

or, $\angle$AOD = ${360}^{\circ }$ - ${221}^{\circ }$ = ${139}^{\circ }$