in a quadrilateral ABCD<A=72degree and <c is the supplementary of <A.the other two angles are 2x-10and x+4.find the value of x and the measure of all the angles
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angle A+angle C= 180°. ( supplementary angle)
angle (A+B+C+D)=360°
180°+2x-10+x+4=360°
3x+174°=360°
x=360°-174°\3=62°
angle A= 72°
angle B= x+4= 62+4=66=
angle C= 180°-72°=108°
angle D = 2x-10=2×62-10=124-10=114
I hope this will satisfy you..
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angle (A+B+C+D)=360°
180°+2x-10+x+4=360°
3x+174°=360°
x=360°-174°\3=62°
angle A= 72°
angle B= x+4= 62+4=66=
angle C= 180°-72°=108°
angle D = 2x-10=2×62-10=124-10=114
I hope this will satisfy you..
PLZZ mark my answer as brainliest
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