Answer
$m\angle A = 78^{\circ}$
$m\angle B = 102^{\circ}$
$m\angle C = 78^{\circ}$
$m\angle D = 102^{\circ}$
Work Step by Step
Opposite angles in a parallelogram have equal measurements, which means ~ $m\angle A$ = $m\angle C$
~ $m\angle B$ = $m\angle D$.
The sum of the interior angles in a parallelogram is $360^{\circ}$, which means if we add two angles that are next to each other, they will be equal to $180^{\circ}$.
1) Two equations can be made out of the information above.
$m\angle A$ and $m\angle B$ are next to each other so $m\angle A$ + $m\angle B$ = $180^{\circ}$
if
$m\angle A$ = 2x + y
$m\angle B$ = 2x + 3y - 20
then
(2x + y) + (2x + 3y -20) = 180
4x + 4y - 20 = 180
4x + 4y - 200 = 0 -- THIS IS THE FIRST EQUATION ! ! !
$m\angle C$ and $m\angle D$ are also next to each other and have not been used so $m\angle C$ + $m\angle D$ = $180^{\circ}$
if
$m\angle C$ = 3x - y + 16
$m\angle D$ = 2x + 3y - 20
then
(3x - y + 16 ) + (2x + 3y - 20) = 180
5x + 2y - 4 = 180
5x + 2y - 184 = 0 -- THIS IS THE SECOND EQUATION ! ! !
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USING THE METHOD OF ELIMINATION, WE CAN FIND OUT THE VALUE OF ONE VARIABLE
2) Multiply the second equation by 2 to make the co-efficient of 'y' in both equations equal to four
4x + 4y - 200 = 0
5x + 2y - 184 = 0 /x2 -- 10x + 4y - 368 = 0
3) Eliminate the variable 'y' by subtracting one equation from another
4x + 4y - 200 = 0
10x + 4y - 368 = 0
(4x + 4y - 200 = 0)
- (10x + 4y - 368 = 0 )
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-6x + 168 + 0
4) Find the value of 'x' .
-6x + 168 + 0
168 = 6x
28 = x
5) Substitute in '28 = x' into an equation to find out the value of 'y'.
[This example will be using the equation ''4x + 4y - 200 = 0'']
4x + 4y - 200 = 0
4(28) + 4y - 200 = 0
112 + 4y - 200 = 0
4y = 0 - 112 + 200
4y = 88
y = 88 $\div$ 4
y = 22
6) Substitute the values of 'x' and 'y' into the expression for each of the angle.
$m\angle A$ = 2x + y
$m\angle A$ = 2(28) + (22)
$m\angle A$ = 56 + 22
$m\angle A$ = 78
$m\angle B$ = 2x + 3y - 20
$m\angle B$ = 2(28) + 3 (22) - 20
$m\angle B$ = 56 + 66 - 20
$m\angle B$ = 102
$m\angle C$ = 3x - y + 16
$m\angle C$ = 3(28) - (22) + 16
$m\angle C$ = 84 - 22 + 16
$m\angle C$ = 78
$m\angle D$ = 2x + 3y - 20
$m\angle D$ = 2(28) + 3 (22) - 20
$m\angle D$ = 56 + 66 - 20
$m\angle D$ = 102
THEREFORE
$m\angle A = 78^{\circ}$
$m\angle B = 102^{\circ}$
$m\angle C = 78^{\circ}$
$m\angle D = 102^{\circ}$
