Question

in a parallelogram ABCD, if angle A= (3x+12)° l, angle B= (2x-32)°, then find the value of x and measure of angle C and ​

Answer 1

Step-by-step explanation:

ABCD is parallelogram

<A&<B are complementary angles

therefore <A+<B=90°

3x+12°+2x-32°=90°

5x-20=90

5x=90+20

5x=110

x=22

<A=<c

3x+12°

3x22+12°

78

<c=78°

Answer 2

$\Large{\bf{\green{\mathfrak{\dag{\underline{\underline{Given:-}}}}}}}$

• Parallelogram ABCD, $\sf \angle A$ = (3x + 12)°, $\sf \angle B$ = (2x - 32)°

$\Large{\bf{\orange{\mathfrak{\dag{\underline{\underline{To \: Find:-}}}}}}}$

• Value of x, measure of $\sf \angle C \: and \: \angle D$

$\Large{\bf{\red{\mathfrak{\dag{\underline{\underline{Solution:-}}}}}}}$

• What do we know about a parallelogram?

• We know that,

Opposite angles are equal.

Adjacent angles add upto 180°.

• Here, $\sf \angle A \: and \: \angle B$ are adjacent angles.

Therefore,

• $\sf \angle A \: + \: \angle B \: = \: 180^{\circ}$

Substituting the values,

• (3x + 12)° + (2x - 32)° = 180°

• 3x° + 12° + 2x° - 32° = 180°

• 5x° - 20° = 180°

• 5x° = 180° + 20°

• 5x° = 200°

• $\sf x^{\circ} \: = \: \dfrac{200^{\circ}}{5}$

• $\sf x^{\circ} \: = \: \dfrac{\cancel{200^{\circ}}}{\cancel{5}}$

• $\boxed{\pink{\sf x^{\circ} \: = \: 40^{\circ}}}$

• Now, we have found out the value of x°.

• Now,

• Let's find the values of $\sf \angle A \: and \: \angle B$

$\sf \angle A$ =

• (3x + 12)°

• (3 * 40 + 12)°

• 120° + 12°

• 132°

$\sf \angle B$ =

• (2x - 32)°

• (2 * 40 - 32)°

• 80° - 32°

• 48°

We know that,

• $\sf \angle A \: = \: \angle C$ (Opposite angles)

• $\sf \angle A \: = \: \angle C$ = 132°

• $\sf \angle B \: = \: \angle D$ (Opposite angles)

• $\sf \angle B \: = \: \angle D$ = 48°

Therefore,

• x° = 40°

• $\sf \angle C$ = 132°

• $\sf \angle D$ = 48°