Mathematics

Two angles of a triangle are equal and the third angle is greater than each one of them by $$18^{\circ}$$. Find the angles.


SOLUTION
Consider $$\angle A$$ and $$\angle B$$ in a triangle is $$x^{\circ}$$

We know that the sum of all the angles in a triangle is $$180^{\circ}$$.

So we can write it as

$$\angle A + \angle B + \angle C = 180^{\circ}$$

By substituting the values

$$x^{\circ} + x^{\circ} + \angle C = 180^{\circ}$$

By addition

$$2x^{\circ} + \angle C = 180^{\circ} .. (1)$$

According to the question we get

$$\angle C = x^{\circ} + 18^{\circ} . (2)$$

By substituting (2) in (1) we get

$$2x^{\circ} + x^{\circ} + 18^{\circ} = 180^{\circ}$$

On further calculation

$$3x^{\circ} + 18^{\circ} = 180^{\circ}$$

By subtraction

$$3x^{\circ} = 180^{\circ} - 18^{\circ}$$

$$3x^{\circ} = 162^{\circ}$$

By division

$$x^{\circ} = 162/3$$

$$x^{\circ} = 54^{\circ}$$

By substituting the values of $$x$$

$$\angle A = \angle B = 54^{\circ}$$

$$\angle C = 54^{\circ} + 18^{\circ} = 72^{\circ}$$

Therefore, $$\angle A = 54^{\circ}, \angle B = 54^{\circ}$$ and $$\angle C = 72^{\circ}$$.
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