Let $OA, OB, OC$ and $OD$ be rays in the anticlockwise direction such that $\angle AOB = 100^{\circ}, \angle COD = 100^{\circ}, \angle BOC = 82^{\circ}$ and $\angle AOD = 78^{\circ}$. Is it true to say that $AOC$ and $BOD$ are lines?

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(NONE) line is formed if the sum of adjacent angles around a point is $180^{\circ}$.
For $AOC$ to be a line,the sum of angles $\angle AOB + \angle BOC$ must be $180^{\circ}$.
Given $\angle AOB = 100^{\circ}$ and $\angle BOC = 82^{\circ}$,their sum is $100^{\circ} + 82^{\circ} = 182^{\circ}$.
Since $182^{\circ} \neq 180^{\circ}$,$AOC$ is not a straight line.
Similarly,for $BOD$ to be a line,the sum of angles $\angle BOC + \angle COD$ must be $180^{\circ}$.
Given $\angle BOC = 82^{\circ}$ and $\angle COD = 100^{\circ}$,their sum is $82^{\circ} + 100^{\circ} = 182^{\circ}$.
Since $182^{\circ} \neq 180^{\circ}$,$BOD$ is also not a straight line.

Explore More

Similar Questions

In the figure,lines $PQ$ and $RS$ intersect each other at point $O$. If $\angle POR : \angle ROQ = 5 : 7,$ find all the angles.

In the figure,$OP$,$OQ$,$OR$,and $OS$ are four rays. Prove that $\angle POQ + \angle QOR + \angle SOR + \angle POS = 360^o$.

An angle whose measure is $90^{\circ}$ is called a $\ldots \ldots \ldots$

$A$ part [or portion] of a line with two end points is called a $\ldots \ldots \ldots$

$A$ straight angle has measure......... (in $^{\circ}$)

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo