Question

The sum of two angles of a triangle is ${115}^{°}$ and their difference is ${45}^{°}$ . Find the measure of each angle.

Answer 1

Let the angles of a triangle are $\angle A,\angle B,\angle C$

then, it is given that, sum of two angles of a triangle is ${115}^{°}$

i.e- $\angle A+\angle B=115$…………………………………$eq1$

Also it is given that- $\angle A-\angle B=45$ ………………………………………$eq2$

Adding both the equations

We have $$\begin{gathered} \angle A+\angle B=115 \\ \angle A-\angle B=45 \\ \Rightarrow 2\angle A=160 \\ \Rightarrow \angle A=\frac{160}{2} \\ \Rightarrow \angle A=80 \end{gathered}$$

We have, $\angle A={60}^{°}$

Then we have to find the value of $\angle B$

So, from $eq1$ it is given that

$\angle A+\angle B=115$

i.e-

$$\begin{gathered} 60+\angle B=115 \\ \angle B=115-60 \\ \angle B=55 \end{gathered}$$

So, we get the value of $\angle B={55}^{°}$

Now, We know that sum of all interior angles of a triangle are ${180}^{°}$

So, we have,

$$\begin{gathered} \angle A+\angle B+\angle C=180 \\ 60+55+\angle C=180 \\ \Rightarrow \angle C+115=180 \\ \Rightarrow \angle C=180-115 \\ \Rightarrow \angle C=65 \end{gathered}$$

So, $\angle C={65}^{°}$

Hence, we have the all three angles are $\angle A={60}^{°}$ , $\angle B={55}^{°}$ and $\angle C={65}^{°}$