Two angles of a triangle are equal and the third angle is greater than

1.	Two angles of a triangle are equal and the third angle is greater than each one of them by 18°. Find the angles.
Answer» Consider ∠A and ∠B in a triangle is x^o We know that the sum of all the angles in a triangle is 180^o. So we can write it as ∠A + ∠B + ∠C = 180^o By substituting the values x^o + x^o + ∠C = 180^o By addition 2x^o + ∠C = 180^o ….. (1) According to the question we get ∠C = x^o + 18^o ……. (2) By substituting (2) in (1) we get 2x^o + x^o + 18^o = 180^o On further calculation 3x^o + 18^o = 180^o By subtraction 3x^o= 180^o – 18^o 3x^o= 162^o By division x^o = 162/3 x^o = 54^o By substituting the values of x ∠A = ∠B = 54^o ∠C = 54^o + 18^o = 72^o Therefore, ∠A = 54^o, ∠B = 54^o and ∠C = 72^o

Two angles of a triangle are equal and the third angle is greater than each one of them by 18°. Find the angles.

Answer»

Consider ∠A and ∠B in a triangle is x^o

We know that the sum of all the angles in a triangle is 180^o.

So we can write it as

∠A + ∠B + ∠C = 180^o

By substituting the values

x^o + x^o + ∠C = 180^o

By addition

2x^o + ∠C = 180^o ….. (1)

According to the question we get

∠C = x^o + 18^o ……. (2)

By substituting (2) in (1) we get

2x^o + x^o + 18^o = 180^o

On further calculation

3x^o + 18^o = 180^o

By subtraction

3x^o= 180^o – 18^o

3x^o= 162^o

By division

x^o = 162/3

x^o = 54^o

By substituting the values of x

∠A = ∠B = 54^o

∠C = 54^o + 18^o = 72^o

Therefore, ∠A = 54^o, ∠B = 54^o and ∠C = 72^o