Chapter 12.4, Problem 5CYU

To determine

To find: Solve $\Delta DEF$ .

Answer to Problem 5CYU

  $\begin{array}{l}d=7.9\\ f=7\\ E={107}^0\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}D={39}^0\\ Q={34}^0\\ e=12\end{array}$

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

  $\begin{array}{l}D={39}^0\\ F={34}^0\\ e=12\end{array}$

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

  $\begin{array}{l}\Rightarrow D+E+F={180}^0\\ \Rightarrow {39}^0+E+{34}^0={180}^0\\ \Rightarrow {73}^0+E={180}^0\\ \Rightarrow {73}^0+E={180}^0-{73}^0\\ \Rightarrow E={180}^0-{73}^0\\ \Rightarrow E={107}^0\end{array}$

Using Law of Sines

We get

  $\frac{\sin \left({39}^0\right)}{d}=\frac{\sin \left({107}^0\right)}{12}=\frac{\sin \left({34}^0\right)}{f}$

Taking first two terms

We get

  $\begin{array}{l}\frac{\sin \left({39}^0\right)}{d}=\frac{\sin \left({107}^0\right)}{12}\\ \Rightarrow d=\frac{12\sin \left({39}^0\right)}{\sin \left({107}^0\right)}\\ \Rightarrow d=\frac{12\left(0.629\right)}{\left(0.9563\right)}\\ \Rightarrow d=\frac{7.548}{\left(0.9563\right)}\\ \Rightarrow d=7.9\end{array}$

Taking Last two terms

We get

  $\begin{array}{l}\frac{\sin \left({107}^0\right)}{12}=\frac{\sin \left({34}^0\right)}{f}\\ \Rightarrow f=\frac{12\sin \left({34}^0\right)}{\sin \left({107}^0\right)}\\ \Rightarrow f=\frac{12\left(0.5592\right)}{\left(0.9563\right)}\\ \Rightarrow f=\frac{6.7104}{\left(0.9563\right)}\\ \Rightarrow f=7\end{array}$

Therefore,

  $\begin{array}{l}d=7.9\\ f=7\\ E={107}^0\end{array}$