Chapter 11.2, Problem 23IP

To determine

To find: Measures of the angles.

Answer to Problem 23IP

Measures of the angles are ${53}^0,{55}^0$ and ${72}^0$

Explanation of Solution

Given information:

Angles of a triangle are $x^0,{\left(x-2\right)}^0$ and ${72}^0$

Given

Angles of a triangle are $x^0,{\left(x-2\right)}^0$ and ${72}^0$

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

Using that

We get

  $x^0+{\left(x-2\right)}^0+{72}^0={180}^0$

On solving

  $\begin{array}{l}{x}^0+{\left(x-2\right)}^0+{72}^0={180}^0\\ \Rightarrow x^0+x^0-2^0+{72}^0={180}^0\\ \Rightarrow 2x^0+{70}^0={180}^0\\ \Rightarrow 2x^0={180}^0-{70}^0\\ \Rightarrow 2x^0={110}^0\\ \Rightarrow x^0=\frac{{110}^0}{2}\\ \Rightarrow x^0={55}^0\end{array}$

Therefore,

  $x={55}^0$

One angle is $x={55}^0$

Second angle is $x-2^0={55}^0-2^0={53}^0$

Third angle is ${72}^0$

Therefore,

Measures of the angles are ${53}^0,{55}^0$ and ${72}^0$