Chapter 12, Problem 12SGR

To determine

To find: Solve $\Delta ABC$ .

Answer to Problem 12SGR

  $\begin{array}{l}b=6.7\\ c=9\\ B={48}^0\\ A={42}^0\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}a=6\\ C={90}^0\\ \sin A=\frac{2}{3}\end{array}$

Concept:

Law of sines

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

Given

  $\begin{array}{l}a=6\\ C={90}^0\\ \sin A=\frac{2}{3}\end{array}$

We know

  $\sin A=\frac{opposite}{hypotenuse}$

On substituting values

We get

  $\begin{array}{l}\Rightarrow \sin A=\frac{a}{c}\\ \Rightarrow \frac{2}{3}=\frac{6}{c}\\ \Rightarrow c=\frac{3\times 6}{2}\\ \Rightarrow c=\frac{18}{2}\\ \Rightarrow c=9\end{array}$

Given

  $\sin A=\frac{2}{3}$

On simplifying

We get

  $\begin{array}{l}\Rightarrow A={\sin }^{-1}\left(\frac{2}{3}\right)\\ \Rightarrow A={42}^0\end{array}$

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

Using that

We get

  $\begin{array}{l}A+B+C={180}^0\\ \Rightarrow {42}^0+B+{90}^0={180}^0\\ \Rightarrow B+{132}^0={180}^0\\ \Rightarrow B={180}^0-{132}^0\\ \Rightarrow B={48}^0\end{array}$

Using law of sines

We get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\\ \Rightarrow \frac{\sin {42}^0}{6}=\frac{\sin {48}^0}{b}=\frac{\sin {90}^0}{9}\end{array}$

Taking last two terms

We get

  $\begin{array}{l}\Rightarrow \frac{\sin {48}^0}{b}=\frac{\sin {90}^0}{9}\\ \Rightarrow b=\frac{9\sin {48}^0}{\sin {90}^0}\\ \Rightarrow b=\frac{9\left(0.743\right)}{1}\\ \Rightarrow b=6.7\end{array}$

Therefore,

  $\begin{array}{l}b=6.7\\ c=9\\ B={48}^0\\ A={42}^0\end{array}$