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

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

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

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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Subjective Medium Published on 09th 09, 2020
Questions 120418
Subjects 10
Chapters 88
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