Chapter 13, Problem 11SGR

To determine

To find: Solve $\Delta ABC$ .

Answer to Problem 11SGR

  $\begin{array}{l}b=18\\ a=12.6\\ B={55}^0\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}c=22\\ A={35}^0\\ C={90}^0\end{array}$

Concept:

Law of sines

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

Given

  $\begin{array}{l}c=22\\ A={35}^0\\ C={90}^0\end{array}$

We know

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

Using that

We get

  $\begin{array}{l}A+B+C={180}^0\\ \Rightarrow {35}^0+B+{90}^0={180}^0\\ \Rightarrow B+{125}^0={180}^0\\ \Rightarrow B={180}^0-{125}^0\\ \Rightarrow B={55}^0\end{array}$

Using law of sines

We get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\\ \Rightarrow \frac{\sin {35}^0}{a}=\frac{\sin {55}^0}{b}=\frac{\sin {90}^0}{22}\end{array}$

Taking last two terms

We get

  $\begin{array}{l}\Rightarrow \frac{\sin {55}^0}{b}=\frac{\sin {90}^0}{22}\\ \Rightarrow b=\frac{22\sin {55}^0}{\sin {90}^0}\\ \Rightarrow b=\frac{22\left(0.82\right)}{1}\\ \Rightarrow b=18\end{array}$

Using law of sines

Taking first and last terms

We get

  $\begin{array}{l}\Rightarrow \frac{\sin {35}^0}{a}=\frac{\sin {90}^0}{22}\\ \Rightarrow a=\frac{22\sin {35}^0}{\sin {90}^0}\\ \Rightarrow a=\frac{22\left(0.5736\right)}{1}\\ \Rightarrow a=12.6\end{array}$

Therefore,

  $\begin{array}{l}b=18\\ a=12.6\\ B={55}^0\end{array}$