Chapter 13.4, Problem 7CYU

To determine

To find: Solve $\Delta FGH$ .

Answer to Problem 7CYU

  $\begin{array}{l}f=12.3\\ h=9.1\\ F={60}^0\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}G={80}^0\\ H={40}^0\\ g=14\end{array}$

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

  $\begin{array}{l}G={80}^0\\ H={40}^0\\ g=14\end{array}$

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

  $\begin{array}{l}\Rightarrow F+G+H={180}^0\\ \Rightarrow F+{80}^0+{40}^0={180}^0\\ \Rightarrow {120}^0+F={180}^0\\ \Rightarrow F={180}^0-{120}^0\\ \Rightarrow F={60}^0\end{array}$

Using Law of Sines

We get

  $\frac{\sin \left({60}^0\right)}{f}=\frac{\sin \left({80}^0\right)}{14}=\frac{\sin \left({40}^0\right)}{h}$

Taking first two terms

We get

  $\begin{array}{l}\frac{\sin \left({60}^0\right)}{f}=\frac{\sin \left({80}^0\right)}{14}\\ \Rightarrow f=\frac{14\sin \left({60}^0\right)}{\sin \left({80}^0\right)}\\ \Rightarrow f=\frac{14\left(0.866\right)}{\left(0.9848\right)}\\ \Rightarrow f=\frac{12.124}{\left(0.9848\right)}\\ \Rightarrow f=12.3\end{array}$

Taking Last two terms

We get

  $\begin{array}{l}\frac{\sin \left({80}^0\right)}{14}=\frac{\sin \left({40}^0\right)}{h}\\ \Rightarrow h=\frac{14\sin \left({40}^0\right)}{\sin \left({80}^0\right)}\\ \Rightarrow h=\frac{14\left(0.6428\right)}{\left(0.9848\right)}\\ \Rightarrow h=\frac{9}{\left(0.9848\right)}\\ \Rightarrow h=9.1\end{array}$

Therefore,

  $\begin{array}{l}f=12.3\\ h=9.1\\ F={60}^0\end{array}$