Chapter 13, Problem 7SGR

To determine

To find:Solve $\Delta ABC$ .

Answer to Problem 7SGR

  $\begin{array}{l}a=10.9\\ B={25}^0\\ C={65}^0\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}c=12\\ b=5\\ C={90}^0\end{array}$

Concept:

Law of sines

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

Given

  $\begin{array}{l}c=12\\ b=5\\ C={90}^0\end{array}$

Using law of sines

We get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\\ \Rightarrow \frac{\sin A}{a}=\frac{\sin B}{5}=\frac{\sin {90}^0}{12}\end{array}$

Taking last two terms

We get

  $\begin{array}{l}\Rightarrow \frac{\sin B}{5}=\frac{\sin {90}^0}{12}\\ \Rightarrow \frac{\sin B}{5}=\frac{1}{12}\\ \Rightarrow \sin B=\frac{5}{12}\\ \Rightarrow \sin B=0.417\\ \Rightarrow B={\sin }^{-1}\left(0.417\right)\\ \Rightarrow B={25}^0\end{array}$

We know

Sum of angle of Triangle is ${180}^0$

Using that

We get

  $\begin{array}{l}A+B+C={180}^0\\ \Rightarrow A+{25}^0+{90}^0={180}^0\\ \Rightarrow A+{115}^0={180}^0\\ \Rightarrow A={180}^0-{115}^0\\ \Rightarrow A={65}^0\end{array}$

Using law of sine

Taking first and last terms

We get

  $\begin{array}{l}\Rightarrow \frac{\sin {65}^0}{a}=\frac{\sin {90}^0}{12}\\ \Rightarrow a=\frac{12\sin {65}^0}{\sin {90}^0}\\ \Rightarrow a=\frac{12\left(0.906\right)}{1}\\ \Rightarrow a=10.9\end{array}$

Therefore,

  $\begin{array}{l}a=10.9\\ B={25}^0\\ C={65}^0\end{array}$