Question

The sum of two angles of a triangle is 116° and their difference is 24°. Find the measure of each angle of the triangle.

Answer 1

Finding the value of $\mathbf{\angle }\mathbf{C}$:

Given data:

In $△\mathrm{ABC}$

1. Sum of two angles is 116°
2. Their difference is 24°

Let us assumes the angles of triangle as A, B, and C

As given above,

$$\begin{gathered} \angle A+\angle B=116° \\ \angle A-\angle B=24° \end{gathered}$$

As we know, that the sum of all angles of a triangle is 180°

The sum of all angles of a triangle is equal to 180°.

So we can write,

$$\begin{gathered} \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180° \\ 116°+\angle \mathrm{C}=180° \\ \angle \mathrm{C}=180°-116° \\ \angle \mathrm{C}=64° \end{gathered}$$

Finding the values of $\mathbf{\angle }\mathbf{A}\mathbf{}\mathbf{and}\mathbf{}\mathbf{\angle }\mathbf{B}\mathbf{}$by substituting and solving it simultaneously:

$$\begin{gathered} \because \angle A+\angle B=24° \\ \therefore \angle A=24°+\angle B·····\left(1\right) \\ and \\ \because \angle A-\angle B=116°·····\left(2\right) \end{gathered}$$

Substituting equation (1) in (2) we get,

$$\begin{gathered} 24°+\angle B+\angle B=116° \\ 2\angle B=116°-24° \\ 2\angle B=92° \\ \angle B=\frac{92°}{2} \\ \angle B=46° \end{gathered}$$

Substituting above value in equation (1) to find $\mathbf{\angle }\mathbf{A}$

$$\begin{gathered} \angle \mathrm{A}=24°+\angle \mathrm{B} \\ \angle \mathrm{A}=24°+46° \\ \angle \mathrm{A}=70° \end{gathered}$$

Hence, the measure of all angles of a triangle are$\mathbf{\angle }\mathbf{A}\mathbf{}\mathbf{=}\mathbf{}\mathbf{70}\mathbf{°}\mathbf{,}\mathbf{}\mathbf{\angle }\mathbf{B}\mathbf{}\mathbf{=}\mathbf{}\mathbf{46}\mathbf{°}\mathbf{,}\mathbf{}\mathbf{\angle }\mathbf{C}\mathbf{}\mathbf{=}\mathbf{}\mathbf{64}\mathbf{°}$.