Question

In a parallelogram ABCD, if angle(A) =(3x+12)°,

Angle(B)= (2x-32)° then find the value of x and angles

C and D.​

Answer 1

$\huge{\bold{\underline{\underline{Question:-}}}}$

In a parallelogram ABCD, if angle(A) =(3x+12)°,

Angle(B)= (2x-32)° then find the value of x and angles

C and D.

$\huge{\bold{\underline{\underline{Solution:-}}}}$

ABCD is a parallelogram. [Given] ∴ ∠A + ∠B = 180° [Adjacent angles of a parallelogram are supplementary],

∴ (3x + 12)° + (2x-32)° = 180°

∴ 3x + 12 + 2x – 32 = 180

∴ 5x – 20 = 180

∴ 5x =180+20

∴ 5x = 200

∴ x = 200/5

∴ x = 40

ii. ∠A = (3x + 12)°

= [3(40) + 12]°

= (120 + 12)° = 132°

∠B = (2x – 32)°

= [2(40) – 32]°

= (80 – 32)° = 48°

∴ ∠C = ∠A = 132°

∠D = ∠B = 48° [Opposite angles of a parallelogram] ∴ The value of x is 40, and the measures of ∠C and

∠D are 132° and 48° respectively.

Answer 2

Answer:

value of

X is 40

C is 132°

D is 48°

Step-by-step explanation:

Angle A + angle B = 180°

ATQ:

Value of X =

3x +12 + 2x- 32= 180°

or 5x =180+20

or X =200÷5

or X = 40

Angle C =

(3x40 + 12)

= (120+12)

= 132°

Angle D =

(2x40-32)

= (80-32)

= 48°