Chapter 11.2, Problem 15IP

To determine

To find: Measures of the angles.

Answer to Problem 15IP

Angles are ${30}^0,{60}^0,{90}^0$

Explanation of Solution

Given information:

One exterior angle measures ${120}^0$

Given

One exterior angle measures ${120}^0$

We know that exterior and its adjacent interior angle forms a straight line

It means they are supplementary angles and sum of their measure is ${180}^0$

Let $x$ be the adjacent interior angle and it is an angle of a triangle

Using that

We get

  $\begin{array}{l}x+{120}^0={180}^0\\ \Rightarrow x={180}^0-{120}^0\\ \Rightarrow x={60}^0\end{array}$

We got one of the angles of triangle is ${60}^0$ and given that triangle is a right angle

So, second angle is ${90}^0$

Let third angle be $y$

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

Using that

We get

  ${60}^0+{90}^0+y={180}^0$

On solving

  $\begin{array}{l}{60}^0+{90}^0+y={180}^0\\ \Rightarrow {150}^0+y={180}^0\\ \Rightarrow y={180}^0-{150}^0\\ \Rightarrow y^0={30}^0\end{array}$

Therefore,

Angles are ${30}^0,{60}^0,{90}^0$