Chapter 12, Problem 10SGR

To determine

To find: Solve $\Delta ABC$ .

Answer to Problem 10SGR

  $\begin{array}{l}b=11.3\\ a=11.3\\ A={45}^0\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}c=16\\ B={45}^0\\ C={90}^0\end{array}$

Concept:

Law of sines

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

Given

  $\begin{array}{l}c=16\\ B={45}^0\\ C={90}^0\end{array}$

We know

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

Using that

We get

  $\begin{array}{l}A+B+C={180}^0\\ \Rightarrow A+{45}^0+{90}^0={180}^0\\ \Rightarrow A+{135}^0={180}^0\\ \Rightarrow A={180}^0-{135}^0\\ \Rightarrow A={45}^0\end{array}$

Using law of sines

We get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\\ \Rightarrow \frac{\sin {45}^0}{a}=\frac{\sin {45}^0}{b}=\frac{\sin {90}^0}{16}\end{array}$

Taking last two terms

We get

  $\begin{array}{l}\Rightarrow \frac{\sin {45}^0}{b}=\frac{\sin {90}^0}{16}\\ \Rightarrow b=\frac{16\sin {45}^0}{\sin {90}^0}\\ \Rightarrow b=\frac{16\left(0.7071\right)}{1}\\ \Rightarrow b=11.3\end{array}$

Using law of sines

Taking first and last terms

We get

  $\begin{array}{l}\Rightarrow \frac{\sin {45}^0}{a}=\frac{\sin {90}^0}{16}\\ \Rightarrow a=\frac{16\sin {45}^0}{\sin {90}^0}\\ \Rightarrow a=\frac{16\left(0.7071\right)}{1}\\ \Rightarrow a=11.3\end{array}$

Therefore,

  $\begin{array}{l}b=11.3\\ a=11.3\\ A={45}^0\end{array}$