In quadrilateral abcd angle a+angle c=140 abgle a is to angle c = 1:3
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Let angle a be x and angle c be 3x
\_a+\_b=140°
x + 3x = 140
4x=140
x=140/4
x=35
therefore, 3x=35×3 = 105
so, angle a=35° and angle c=105°.
\_a+\_b=140°
x + 3x = 140
4x=140
x=140/4
x=35
therefore, 3x=35×3 = 105
so, angle a=35° and angle c=105°.
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