Chapter 9.6, Problem 1Q

Solution

To solve: the $△ABC$ .

  $C=56°,a\approx 11.09,b\approx 13.91$

Given information:

  $A=50°,B=74°,c=12$

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,

  $\begin{array}{c}A+B+C=180°\\ 50°+74°+C=180°\\ C=180°-50°-74°\\ C=56°\end{array}$

Now, by the law of Sines,

  $\begin{array}{l}\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\\ \frac{a}{\sin 50°}=\frac{b}{\sin 74°}=\frac{12}{\sin 56°}\end{array}$

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

  $\begin{array}{c}\frac{a}{\sin 50°}=\frac{12}{\sin 56°}\\ a=\frac{12\sin 50°}{\sin 56°}\\ a\approx 11.09\end{array}$

The second equation is:

  $\begin{array}{c}\frac{b}{\sin 74°}=\frac{12}{\sin 56°}\\ b=\frac{12\sin 74°}{\sin 56°}\\ b\approx 13.91\end{array}$