Question

Find the measure of all the angles of a parallelogram if one angle isc24°less than the twice of the smallest angle​

Answer 1

Answer:

hope this helps and pls mark me as brainliest..according to me there are two ways....

Step-by-step explanation:

1st method

Let smaller angle be x Then 2nd angle = 2x -24 In parallelogram, The angle opposite to each other are equal and sum of all angles= 360 Then. - ( 2x -24)+(2x-24)+x + x= 360 2x + 2x + x + x- 48= 360 6x-48=360 6x=360+48 6x=408 X=408÷6 X= 68 And the other = (68×2)- 24= 112

2nd method

In the parallelogram ABCD

                        angle A+angle B= angle C + angle D= 180

                  Let angle A=x

                  then angle B= 2x-24

          We have x+2x-24=180

                                  3x= 204 ⇒x=68         

                        Angle B= 68*2-24=112

Hence the angles are 68,112,68 and 112.

Answer 2

Answer:

angles are 112,68,112,68

Step-by-step explanation:

let the angles be 2x-24,x

$2x - 24 + x = 180 \\ 3x - 24 = 180 \\ 3x = 180 + 24 \\ 3x = 204 \\ x = \frac{204}{3} \\ x = 68 \\ 2x - 24 = 2(68) - 24 \\ 136 - 24 = 112$