question_answer

AB is a straight line and O is a point lying on AB. A line OC is drawn from O such that \[\angle COA=36{}^\circ .\] OD is a line within the \[\angle COA\]such that \[\angle DOA=\left( \frac{1}{3} \right)\,\,\angle COA.\] If OE is a line within the \[\angle BOC,\] \[\angle BOE=\left( \frac{1}{4} \right)\,\,\angle BOC,\] then the \[\angle DOE\] must be

A)  \[60{}^\circ \]

B)  \[45{}^\circ \]  

C)  \[36{}^\circ \] 

D)  \[80{}^\circ \]

Correct Answer: A

Solution :

[a] \[\angle DOA=\frac{1}{3}\angle COA\] and \[\angle BOE=\frac{1}{4}\angle BOC\]

\[\therefore \]\[\angle DOC=\frac{2}{3}\angle COA\] and \[\angle COE=\frac{3}{4}\,\angle BOC\]

\[\Rightarrow \]   \[\angle DOC=\frac{2}{3}\times 36{}^\circ =24{}^\circ \]

and \[\angle COE=\frac{1}{4}(180{}^\circ -36{}^\circ )=\frac{1}{4}\times 144{}^\circ =36{}^\circ \]

\[\therefore \]\[\angle DOE=\angle DOC+\angle COE=24{}^\circ +36{}^\circ =60{}^\circ \]