Two supplementary angles ( x + 42°) and ( 3x - 62°) find x.
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(x+42°)+(3x-62°)=180
4x-20=180
4x=180+20
4x=200
x=200/4=50
x+42=50+42=72°
3x-62=3*50-62=150-62=88°
4x-20=180
4x=180+20
4x=200
x=200/4=50
x+42=50+42=72°
3x-62=3*50-62=150-62=88°
sasir2000:
sry
the second angle is 98
50+42=92
150-62=88
oops sry
i was in a hury
*hurry
i wont give wrong answers anymore
Ok
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