Find the measure of each angle of a parallelogram, if one of its angle

1.	Find the measure of each angle of a parallelogram, if one of its angles is 30° less than twice the smallest angle.
Answer» Consider ABCD as a parallelogram Let us take ∠ A as the smallest angle So we get ∠ B = 2 ∠ A – 30^o We know that the opposite angles are equal in a parallelogram ∠ A = ∠ C and ∠ B = ∠ D = 2 ∠ A – 30^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 ∠ A + (2 ∠ A – 30^o) + ∠ A + (2 ∠ A – 30^o) = 360^o On further calculation ∠ A + 2 ∠ A – 30^o + ∠ A + 2 ∠ A – 30^o = 360^o So we get 6 ∠ A – 60^o = 360^o By addition 6 ∠ A = 360^o + 60^o 6 ∠ A = 420^o By division ∠ A = 70^o By substituting the value of ∠ A ∠ A = ∠ C = 70^o ∠ B = ∠ D = 2 ∠ A – 30^o = 2 (70^o) – 30^o ∠ B = ∠ D = 110^o Therefore, ∠ A = ∠ C = 70^o and ∠ B = ∠ D = 110^o.

Find the measure of each angle of a parallelogram, if one of its angles is 30° less than twice the smallest angle.

Answer»

Consider ABCD as a parallelogram

Let us take ∠ A as the smallest angle

So we get

∠ B = 2 ∠ A – 30^o

We know that the opposite angles are equal in a parallelogram

∠ A = ∠ C and ∠ B = ∠ D = 2 ∠ A – 30^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

∠ A + (2 ∠ A – 30^o) + ∠ A + (2 ∠ A – 30^o) = 360^o

On further calculation

∠ A + 2 ∠ A – 30^o + ∠ A + 2 ∠ A – 30^o = 360^o

So we get

6 ∠ A – 60^o = 360^o

By addition

6 ∠ A = 360^o + 60^o

6 ∠ A = 420^o

By division

∠ A = 70^o

By substituting the value of ∠ A

∠ A = ∠ C = 70^o

∠ B = ∠ D = 2 ∠ A – 30^o = 2 (70^o) – 30^o

∠ B = ∠ D = 110^o

Therefore, ∠ A = ∠ C = 70^o and ∠ B = ∠ D = 110^o.