Chapter 11.2, Problem 22IP

To determine

To find: Measures of the angles.

Answer to Problem 22IP

Measures of the angles are ${45}^0,{45}^0$ and ${90}^0$

Explanation of Solution

Given information:

Angles of a triangle are $x^0,x^0$ and $2x^0$

Given

Angles of a triangle are $x^0,x^0$ and $2x^0$

We know sum of angles of a triangle is ${180}^0$

Using that

We get

  $x^0+x^0+2x^0={180}^0$

On solving

  $\begin{array}{l}{x}^0+x^0+2x^0={180}^0\\ \Rightarrow 4x^0={180}^0\\ \Rightarrow x^0=\frac{{180}^0}{4}\\ \Rightarrow x^0={45}^0\end{array}$

Therefore,

  $x={45}^0$

One angle is $x={45}^0$

Second angle is $x={45}^0$

Third angle is $2x=2\left({45}^0\right)={90}^0$

Therefore,

Measures of the angles are ${45}^0,{45}^0$ and ${90}^0$