Question

1) In a parallelogram ABCD, If ZA ( 3x + 12), ZB = (2x - 32)º then find the value ofx and then find the
measures of ZC and ZD.​

Answer 1

Answer:

c - 128

d - 52

this is the answer

Answer 2

$\Uparrow$See the attachment.

$\huge \bigstar$$\huge\bold{\mathtt{\purple{✍︎A{\pink{N{\green{S{\blue{W{\red{E{\orange{R✍︎}}}}}}}}}}}}}$$\huge \Rightarrow$

$\large\bold \red{Solution:}\Downarrow$

$\Box ABCD$ is a parallelogram.

∴$\angle A \angle B=180°$ ...(Adjacent

angles of parallelogram are supplementary.)

$∴(3x+12)°+(2x-32)°=180°$

$∴ 3x+12+2x-32=180$

$∴5x-20=180$

$∴5x=180+20$

$∴5x=200$

$\large ∴x=\frac{200}{5}$

$∴x =40°$

$\angle C = \angle A$ .(Opposite angles of a parallelogram)

$∴\angle C=3x+12$

$∴\angle C=3(40)+12$

$∴\angle C=120+12$

$∴\angle C=132°$

$∴\angle D=\angle B$ ...(Opposite angles of a parallelogram)

$∴ \angle D=2x-32$

$∴ \angle D= 2(40)-32$

$∴ \angle D= 80-32$

$∴ \angle D=48°$

$\Large\bold \purple{Ans:}\rightarrow$

$x=40°,\angle C = 132°,\angle D=48°.$