Chapter 12.4, Problem 23PPS

To determine

To find: Solve $\Delta LMN$ .

Answer to Problem 23PPS

  $\begin{array}{l}m=4.9\\ n=3.1\\ N={74}^0\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}M={70}^0\\ N={36}^0\\ l=5\end{array}$

Concept:

Sum of angles in a Triangle $={180}^0$

Law of Sines:

  $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

Given

  $\begin{array}{l}M={70}^0\\ N={36}^0\\ l=5\end{array}$

Sum of angles in a Triangle $={180}^0$

  $\begin{array}{l}\Rightarrow L+M+N={180}^0\\ \Rightarrow {70}^0+{36}^0+N={180}^0\\ \Rightarrow {106}^0+N={180}^0\\ \Rightarrow N={180}^0-{106}^0\\ \Rightarrow N={74}^0\end{array}$

Using Law of Sines

We get

  $\frac{\sin \left({74}^0\right)}{5}=\frac{\sin \left({70}^0\right)}{m}=\frac{\sin \left({36}^0\right)}{n}$

Taking first two terms

We get

  $\begin{array}{l}\frac{\sin \left({74}^0\right)}{5}=\frac{\sin \left({70}^0\right)}{m}\\ \Rightarrow m=\frac{5\sin \left({70}^0\right)}{\sin \left({74}^0\right)}\\ \Rightarrow m=\frac{5\left(0.9396\right)}{\left(0.9613\right)}\\ \Rightarrow m=\frac{4.698}{\left(0.9613\right)}\\ \Rightarrow m=4.9\end{array}$

Taking First and Last terms

We get

  $\begin{array}{l}\frac{\sin \left({74}^0\right)}{5}=\frac{\sin \left({36}^0\right)}{n}\\ \Rightarrow n=\frac{5\sin \left({36}^0\right)}{\sin \left({74}^0\right)}\\ \Rightarrow n=\frac{5\left(0.5878\right)}{\left(0.9613\right)}\\ \Rightarrow n=\frac{2.939}{\left(0.9613\right)}\\ \Rightarrow n=3.1\end{array}$

Therefore,

  $\begin{array}{l}m=4.9\\ n=3.1\\ N={74}^0\end{array}$