If an angle of a parallelogram is two-thirds of its adjacent angle, find the angles of the parallelogram.

Let a parallelogram be ABCD.

Let ∠A = x and so adjacent angle (∠B) = $\dfrac{2}{3}x$

As AD || BC, sum of co-int ∠s = 180°,

$\therefore x + \dfrac{2}{3}x = 180° \\[1em] \Rightarrow \dfrac{3x + 2x}{3} = 180° \\[1em] \Rightarrow \dfrac{5x}{3} = 180° \\[1em] \Rightarrow x = \dfrac{180° \times 3}{5} \\[1em] \Rightarrow x = 108° \\[1em] \Rightarrow \dfrac{2x}{3} = \dfrac{2 \times 108°}{3} = 72°.$

Opposite angles of a parallelogram are equal.

∴ ∠C = ∠A = 108° and ∠D = ∠B = 72°

Hence, angles of parallelogram are 108°, 72°, 108° and 72°.