in the figure, find the four angles A , B , C and D of the parallelogram ABCD ( answer = 90 ° each ) class 8th .
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in angle BCD
∆DBC+∆C+∆CDB= 180 °
3x+ 5x+2x= 180°
10x= 180°
x=180/10= 18°
∆C=5x=5*18= 90°
now we know that alternate angles are equal it mns
angle DBC= angle ADB= 3x =3*18=54°
angle BDC= angle ABD= 2x=2*18=36°
now , we know
angle DBC + angle ABD= angle B
3x+2x= 5x= 5*18= 90°
angle B= 90°
same for angle D
angle ADB+ angle CDB= angle D
3x+2x= 5x= 5*18=90°
angle D=90°
now in ∆ABD( sum of triangle is 180°)
angle A+ angle ABD+ angle BDA= 180°
angle A+ 36°+ 54°= 180°
angle A +90°= 180°
angle A= 180 -90 = 90°
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