Question

At the intersection of two lines, sum of one pair of opposite angles is ${120}^{\circ }$ more than the sum of the other pair of opposite angles. Find all the angles.

• A. ${30}^{\circ }$, ${30}^{\circ }$, ${150}^{\circ }$, ${150}^{\circ }$
• B. ${60}^{\circ }$, ${60}^{\circ }$, ${120}^{\circ }$, ${120}^{\circ }$
• C. ${30}^{\circ }$, ${30}^{\circ }$, ${60}^{\circ }$, ${60}^{\circ }$
• D. ${15}^{\circ }$, ${15}^{\circ }$, ${75}^{\circ }$, ${75}^{\circ }$

Answer 1

The correct option is B

${60}^{\circ }$, ${60}^{\circ }$, ${120}^{\circ }$, ${120}^{\circ }$

Let sum of one pair of opposite angles be 'x', then sum of the other pair of opposite angles will be ${120}^{\circ }$ + x

Sum of all the four angles at the intersection of two lines = ${360}^{\circ }$

So, x + (x + ${120}^{\circ }$) = ${360}^{\circ }$

or x = ${120}^{\circ }$

Sum of the other pair of opposite angles will be, ${120}^{\circ }$ + ${120}^{\circ }$ = ${240}^{\circ }$

Since oppposite angles are equal, each angle of first pair will be $\frac{{120}^{\circ }}{2}$ = ${60}^{\circ }$

Similarly, each angle of the other pair will be $\frac{{240}^{\circ }}{2}$ = ${120}^{\circ }$