Given parallelogram ABCD, find x, y, and z.
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\(z-15°+ 2z=180°\\ 3z=165°\)
\(z=55°\)
\(x=90°+(90°-(z-15°))\\ x=90°+(90°-(55°-15°))\)
\(x=140°\)
\(y=90°-(2z-90°)\\ y=90°-(2\cdot 55°-90°) \)
\(y=70°\)
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last step should be y = 40
(z - 15) + 2z = 180
3z = 195
z = 65
x = 2z = 130
y = z - 15 = 50
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