Question

AB parallel CD in trapezium ABCD if angle A = Y + 60° and Angle B = X + 60° , angle C=3x-40° and angle D =3y-80°, then find all the angles of ABCD.
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please help me​

Answer 1

Answer:

110,100,80,70 are the values of angles A,B,C,D respectively

Step-by-step explanation:

A trapezium is a quadrilateral with one pair of opposite sides parallel.

Using co-interior angles, we can see that a trapezium has two pairs of adjacent supplementary angles.

so

$\angle A+\angle D=y+60+3y-80=180\\y+60+3y-80=180\\4y-20=180\\4y=200\\y=50$

Similarly

$\angle \:B+\angle \:C=x+60+3x-40=180\\x+60+3x-40=180\\4x+20=180\\x=40$

Now we know y=50 and x=40

we can fin values of all angles by substitution

$\angle \:A=50+60=110\\\angle \:B=40+60=100\\\angle \:C=3\cdot 40-40=80\\\angle \:D=3\cdot 50-80=70$

Answer 2

Answer:

∠A = 110°

∠B = 100°

∠C = 80°

∠D = 70°

Step-by-step explanation:

AB parallel CD in trapezium ABCD if angle A = Y + 60° and Angle B = X + 60° , angle C=3x-40° and angle D =3y-80°, then find all the angles of ABCD.

A = Y + 60°

B = X + 60°

C=3x-40°

D =3y-80°

Sum of all angles = 360°

Y + 60° + X + 60° + 3x-40° + 3y-80° = 360°

4X + 4Y = 360°

X + Y = 90°

AB ║ CD

∠A + ∠D = 180°

∠B + ∠C = 180°

∠A + ∠D = 180°

Y + 60° +  3Y -80° = 180°

4Y = 200°

Y = 50°

X + 50° = 90°

X = 40°

∠A = Y + 60° = 50° + 60° = 110°

∠B = X + 60° = 40° + 60° = 100°

∠C = 3X - 40° = 3*40° - 40° = 80°

∠D = 3y - 80° = 3*50° - 80° = 70°