If an angle of a parallelogram is four fifths of its adjacent angle, f

1.	If an angle of a parallelogram is four fifths of its adjacent angle, find the angles of the parallelogram.
Answer» Consider ABCD as a parallelogram If ∠ A = x^o We know that ∠ B is adjacent to A which can be written as 4/5 x^o Opposite angles are equal in a parallelogram So we get ∠ A = ∠ C = x^o and ∠ B = ∠ D = 4/5 x^o We know that the sum of all the angles of a parallelogram is 360^o It can be written as ∠ A + ∠ B + ∠ C + ∠ D = 360^o By substituting the values in the above equation x + (4/5) x + x + (4/5) x = 360^o By addition we get 2x + (8/5) x = 360^o By taking the LCM as 5 (18/5) x = 360^o By cross multiplication x = (360 × 5)/18 On further calculation x = 100^o By substituting the value of x So we get ∠ A = ∠ C = x = 100^o ∠ B = ∠ D = 4/5 x^o = (4/5) (100^o) = 80^o Therefore, ∠ A = ∠ C = x = 100^o and ∠ B = ∠ D = 80^o.

If an angle of a parallelogram is four fifths of its adjacent angle, find the angles of the parallelogram.

Answer»

Consider ABCD as a parallelogram

If ∠ A = x^o

We know that ∠ B is adjacent to A which can be written as 4/5 x^o

Opposite angles are equal in a parallelogram

So we get

∠ A = ∠ C = x^o and ∠ B = ∠ D = 4/5 x^o

We know that the sum of all the angles of a parallelogram is 360^o

It can be written as

∠ A + ∠ B + ∠ C + ∠ D = 360^o

By substituting the values in the above equation

x + (4/5) x + x + (4/5) x = 360^o

By addition we get

2x + (8/5) x = 360^o

By taking the LCM as 5

(18/5) x = 360^o

By cross multiplication

x = (360 × 5)/18

On further calculation

x = 100^o

By substituting the value of x

So we get

∠ A = ∠ C = x = 100^o

∠ B = ∠ D = 4/5 x^o = (4/5) (100^o) = 80^o

Therefore, ∠ A = ∠ C = x = 100^o and ∠ B = ∠ D = 80^o.

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