Chapter 15.2, Problem 11PSB

(a)

To determine

To Find: The angles of triangles in order beginning with smallest one.

Answer to Problem 11PSB

  $\begin{array}{l}beginning\text{ with the smallest }\\ \angle R<\angle Q<\angle P\end{array}$

Explanation of Solution

Given:

  

Formula Used:

Angle-Sum theorem: Sum of interior angle of a triangle is 180.

Calculation:

  $\begin{array}{l}\text{QR being the largest side}\\ \angle \text{P is the largest of all}\\ \text{PR being the middle side}\\ \angle \text{Q is the inbetween angle}\\ \text{PQ is the smallest side}\\ \angle \text{R is the smallest of all}\text{.}\\ \angle R<\angle Q<\angle P\end{array}$

Conclusion:

Thus, $\angle R<\angle Q<\angle P$

(b)

To determine

To Find: The largest exterior angle.

Answer to Problem 11PSB

  $\angle R\text{ has the largest exterior angle}\text{. }$

Explanation of Solution

Given:

  

Formula Used:

Angle-Sum theorem: Sum of interior angle of a triangle is 180.

Calculation:

  $\begin{array}{l}\text{since the smallest angle is }\angle R\\ \text{it has the largest exterior angle}\text{.}\end{array}$

Conclusion:

Thus, $\angle R\text{ has the largest exterior angle}\text{. }$