Mathematics

In $$\triangle ABC$$, if $$\angle A + \angle B = 125^{\circ}$$ and $$\angle A + \angle C = 113^{\circ}$$, find $$\angle A, \angle B$$ and $$\angle C$$.


SOLUTION
It is given that $$\angle A + \angle B = 125^{\circ} \dots (1)$$

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 $$\angle A + \angle B = 125^{\circ}$$ in the above equation

$$125^{\circ} + \angle C = 180^{\circ}$$

On further calculation

$$\angle C = 180^{\circ} - 125^{\circ}$$

By subtraction

$$\angle C = 55^{\circ}$$

It is given that $$\angle A + \angle C = 113^{\circ}$$

By substituting the value of $$\angle C$$

$$\angle A + 55^{\circ} = 113^{\circ}$$

On further calculation

$$\angle A = 113^{\circ} - 55^{\circ}$$

By subtraction

$$\angle A = 58^{\circ}$$

By substituting $$\angle A = 58^{\circ}$$ in equation (1)

So we get

$$\angle A + \angle B = 125^{\circ}$$

$$58^{\circ} + \angle B= 125^{\circ}$$

On further calculation

$$\angle B = 125^{\circ} - 58^{\circ}$$

By subtraction

$$\angle B = 67^{\circ}$$

Therefore, $$\angle A = 58^{\circ}, \angle B = 67^{\circ}$$ and $$\angle C = 55^{\circ}$$.
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