Chapter 9, Problem 18REAE

Solution

To solve: the $△ABC$ .

  $B=54°,a\approx 10.12,c\approx 14.72$

Given information:

  $A=43°,C=83°,b=12$

Definition Used:

Law of Sines:

The law of sines can be written in either of the following forms for $△ABC$ with sides of length a , b and c .

  

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\\ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\end{array}$

Explanation:

It is known that the sum of all the angles of a triangle is $180°$ .

Using this knowledge to find angle A,

  $\begin{array}{c}A+B+C=180°\\ 43°+B+83°=180°\\ B=180°-43°-83°\\ B=54°\end{array}$

Now, by the law of Sines,

  $\begin{array}{l}\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\\ \frac{a}{\sin 43°}=\frac{12}{\sin 54°}=\frac{c}{\sin 83°}\end{array}$

Use this to write two equations with one variable each and solve them independently to find the value of the variable.

  $\begin{array}{c}\frac{a}{\sin 43°}=\frac{12}{\sin 54°}\\ a=\frac{12\sin 43°}{\sin 54°}\\ a\approx 10.12\end{array}$

Now, the second equation is:

  $\begin{array}{c}\frac{12}{\sin 54°}=\frac{c}{\sin 83°}\\ c=\frac{12\sin 83°}{\sin 54°}\\ c\approx 14.72\end{array}$