Chapter 9.5, Problem 12E

Solution

To solve: the $△ABC$ .

  $A=48°,b\approx 25.47,c\approx 18.7$

Given information:

The image given is:

  

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 .

  

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

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

Explanation:

Comparing the given triangle with the following figure:

  

  $B=85°,C=47°,a=19$

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

Using this knowledge to find angle A,

  $\begin{array}{c}A+B+C=180°\\ A+85°+47°=180°\\ A=180°-85°-47°\\ A=48°\end{array}$

Now, by the law of Sines,

  $\begin{array}{l}\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\\ \frac{19}{\sin 48°}=\frac{b}{\sin 85°}=\frac{c}{\sin 47°}\end{array}$

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

  $\begin{array}{c}\frac{19}{\sin 48°}=\frac{b}{\sin 85°}\\ b=\frac{19\sin 85°}{\sin 48°}\\ b\approx 25.47\end{array}$

And second equation is:

  $\begin{array}{c}\frac{19}{\sin 48°}=\frac{c}{\sin 47°}\\ c=\frac{19\sin 47°}{\sin 48°}\\ c\approx 18.7\end{array}$