Chapter 12.5, Problem 20PPS

To determine

To find: Solve the given triangle.

Answer to Problem 20PPS

  $\begin{array}{l}N={53}^0\\ m=28.4\\ p=38.2\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}M={47}^0\\ P={80}^0\\ n=31\end{array}$

Given two angles and one side of a triangle.

So, the triangle should be solved by beginning with law of Sines.

Sum of angles of a triangle is ${180}^0$

Using that

We get

  $M+N+P={180}^0$

On substituting values

We get

  $\begin{array}{l}M+N+P={180}^0\\ \Rightarrow {47}^0+{80}^0+N={180}^0\\ \Rightarrow {127}^0+N={180}^0\\ \Rightarrow N={180}^0-{127}^0\\ \Rightarrow N={53}^0\end{array}$

Using law of sines

We get

  $\frac{\sin N}{n}=\frac{\sin M}{m}$

On substituting values

We get

  $\begin{array}{l}\frac{\sin N}{n}=\frac{\sin M}{m}\\ \Rightarrow \frac{\sin {53}^0}{31}=\frac{\sin {47}^0}{m}\\ \Rightarrow m=\frac{31\sin {47}^0}{\sin {53}^0}\\ \Rightarrow m=\frac{31\left(0.7313\right)}{0.7986}\\ \Rightarrow m=\frac{22.67}{0.7986}\\ \Rightarrow m=28.4\end{array}$

Using law of sines

We get

  $\frac{\sin N}{n}=\frac{\sin P}{p}$

On substituting values

We get

  $\begin{array}{l}\frac{\sin N}{n}=\frac{\sin P}{p}\\ \Rightarrow \frac{\sin {53}^0}{31}=\frac{\sin {80}^0}{p}\\ \Rightarrow p=\frac{31\sin {80}^0}{\sin {53}^0}\\ \Rightarrow p=\frac{31\left(0.9848\right)}{0.7986}\\ \Rightarrow p=\frac{30.53}{0.7986}\\ \Rightarrow p=38.2\end{array}$

Therefore,

  $\begin{array}{l}N={53}^0\\ m=28.4\\ p=38.2\end{array}$