In each of the following figures, ABCD is a parallelogram.

(i)

(ii)

In each case, given above, find the values of x and y.

(i) We know that,

Opposite sides of parallelogram are equal.

∴ AB = CD and AD = BC

⇒ AB = CD

⇒ 4x = 6y + 2

⇒ x = $\dfrac{6y + 2}{4}$ ……..(1)

⇒ AD = BC

⇒ 4y = 3x - 3

⇒ 3x = 4y + 3

⇒ x = $\dfrac{4y + 3}{3}$ ……….(2)

From equation (1) and (2), we get :

⇒ $\dfrac{6y + 2}{4} = \dfrac{4y + 3}{3}$

⇒ 3(6y + 2) = 4(4y + 3)

⇒ 18y + 6 = 16y + 12

⇒ 18y - 16y = 12 - 6

⇒ 2y = 6

⇒ y = $\dfrac{6}{2}$ = 3.

Substituting value of y in equation (1), we get :

⇒ x = $\dfrac{6y + 2}{4} = \dfrac{6 \times 3 + 2}{4} = \dfrac{18 + 2}{4} = \dfrac{20}{4}$ = 5.

Hence, x = 5 and y = 3.

(ii) We know that,

Opposite angles of parallelogram are equal.

∴ ∠B = ∠D

⇒ 7y = 6x + 3y - 8°

⇒ 7y - 3y = 6x - 8°

⇒ 4y = 6x - 8°

⇒ y = $\dfrac{6x - 8°}{4}$ ……….(1)

We know that,

Consecutive angles of a parallelogram are supplementary.

⇒ ∠A + ∠C = 180°

⇒ 4x + 20° + 7y = 180°

⇒ 4x + 7y = 180° - 20°

⇒ 4x + 7y = 160°

⇒ 7y = 160° - 4x

⇒ y = $\dfrac{160° - 4x}{7}$ ………..(2)

From equation (1) and (2), we get :

⇒ $\dfrac{6x - 8°}{4} = \dfrac{160° - 4x}{7}$

⇒ 7(6x - 8°) = 4(160° - 4x)

⇒ 42x - 56° = 640° - 16x

⇒ 42x + 16x = 640° + 56°

⇒ 58x = 696°

⇒ x = $\dfrac{696°}{58}$ = 12°.

Substituting value of x in equation (1), we get :

⇒ y = $\dfrac{6x - 8°}{4} = \dfrac{6 \times 12° - 8°}{4} = \dfrac{72° - 8°}{4} = \dfrac{64°}{4}$ = 16°.

Hence, x = 12° and y = 16°.