Question

The sum of two opposite angles of a parallelogram is 120${}^{\circ }$. Find the measure of each of its angles.

• A. 60${}^{\circ }$, 120${}^{\circ }$, 60${}^{\circ }$ and 120.${}^{\circ }$
• B. 30${}^{\circ }$, 150${}^{\circ }$, 30${}^{\circ }$ and 150.${}^{\circ }$
• C. 90${}^{\circ }$, 45${}^{\circ }$, 60${}^{\circ }$ and 120.${}^{\circ }$
• D. 60${}^{\circ }$, 90${}^{\circ }$, 60${}^{\circ }$ and 90.${}^{\circ }$

Answer 1

The correct option is A 60${}^{\circ }$, 120${}^{\circ }$, 60${}^{\circ }$ and 120.${}^{\circ }$


Let, ABCD be a parallelogram.

Given, sum of its opposite angles be equal to 120${}^{\circ }$
$\Rightarrow$ ∠A+∠C = 120${}^{\circ }$

The opposite angles are equal in a parallelogram.
So, ∠A = ∠C = x${}^{\circ }$
$\Rightarrow$ x${}^{\circ }$ + x${}^{\circ }$ = 120${}^{\circ }$
$\Rightarrow$ x${}^{\circ }$ = 60${}^{\circ }$
$\Rightarrow$ ∠A = 60${}^{\circ }$ and ∠C = 60${}^{\circ }$

∠A + ∠B = 180${}^{\circ }$
[$\because$ Sum of adjacent angles of a
parallelogram is 180°]
$\Rightarrow$ 60${}^{\circ }$ + ∠B = 180${}^{\circ }$
$\Rightarrow$ ∠B = (180−60)${}^{\circ }$
$\Rightarrow$ ∠B = 120${}^{\circ }$
$\Rightarrow$ ∠D = ∠B = 120${}^{\circ }$
[Opposite angles of parallelogram are equal]