Question

determine as medidas dos ângulos internos deste paralelogramos

Answer 1

3x + x/2 + 40 = 180
(3x + x/2 + 40)×2 = (180)×2
6x + x + 80 = 360
7x = 360 - 80
7x = 280
x = 280/7
x = 40

3x = z
3(40) = z
120 = z

x/2 + 40 = y
40/2 + 40 = y
(40/2 + 40)×2 = (y)×2
40 + 80 = 2y
120 = 2y
2y = 120
y = 120/2
y = 60