Chapter 13.4, Problem 2CYP

To determine

To find:Solve $\Delta NPQ$ .

Answer to Problem 2CYP

  $\begin{array}{l}p=3.5\\ q=4.74\\ N={73}^0\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}P={42}^0\\ Q={65}^0\\ n=5\end{array}$

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

  $\begin{array}{l}P={42}^0\\ Q={65}^0\\ n=5\end{array}$

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

  $\begin{array}{l}\Rightarrow P+Q+N={180}^0\\ \Rightarrow {42}^0+{65}^0+N={180}^0\\ \Rightarrow {107}^0+N={180}^0\\ \Rightarrow N={180}^0-{107}^0\\ \Rightarrow N={73}^0\end{array}$

Using Law of Sines

We get

  $\frac{\sin \left({42}^0\right)}{p}=\frac{\sin \left({65}^0\right)}{q}=\frac{\sin \left({73}^0\right)}{5}$

Taking last two terms

We get

  $\begin{array}{l}\Rightarrow \frac{\sin \left({65}^0\right)}{q}=\frac{\sin \left({73}^0\right)}{5}\\ \Rightarrow q=\frac{5\times \sin \left({65}^0\right)}{\sin \left({73}^0\right)}\\ \Rightarrow q=\frac{5\times \left(0.9063\right)}{\left(0.9563\right)}\\ \Rightarrow q=\frac{4.5315}{\left(0.9563\right)}\\ \Rightarrow q=4.74\end{array}$

Taking First and Last terms

We get

  $\begin{array}{l}\frac{\sin \left({42}^0\right)}{p}=\frac{\sin \left({73}^0\right)}{5}\\ \Rightarrow p=\frac{5\sin \left({42}^0\right)}{\sin \left({73}^0\right)}\\ \Rightarrow p=\frac{5\left(0.669\right)}{0.9563}\\ \Rightarrow p=\frac{3.345}{0.9563}\\ \Rightarrow p=3.5\end{array}$

Therefore,

  $\begin{array}{l}p=3.5\\ q=4.74\\ N={73}^0\end{array}$