ABCD is a trapezium find X and Y please gave me full explanation step by step
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the answer of your question is in the attachment....
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rajkumar3964:
Your answer is very useful for me
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hey!!!
In triangle APB
P-40°
A-180-100(Supplementary angle)
A-80°
Sum of all sides of a triangle -180°
So A+P+B=180°
80+40+B=180
120+B=180
180-120=B
B=60°
In angle ABC
B=180-60
=120°
Alternate angles of a trapezium add up to 180°
angle BCD=180-120=60°
Y=60°
X=40+60+x=180(sum of triangle PDC)
x=100+x=180
x=180-100
x=80°
X=80°
Y=60°
Hope it helps you
Please mark it as brainliest.
In triangle APB
P-40°
A-180-100(Supplementary angle)
A-80°
Sum of all sides of a triangle -180°
So A+P+B=180°
80+40+B=180
120+B=180
180-120=B
B=60°
In angle ABC
B=180-60
=120°
Alternate angles of a trapezium add up to 180°
angle BCD=180-120=60°
Y=60°
X=40+60+x=180(sum of triangle PDC)
x=100+x=180
x=180-100
x=80°
X=80°
Y=60°
Hope it helps you
Please mark it as brainliest.
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