Chapter 9.6, Problem 2GP

Solution

To solve: the $△ABC$ .

  $A\approx 60.69°,B\approx 85.22°,C\approx 34.09°$

Given information:

The given measurements are:

  $a=14,b=16,c=9$

Definition Used:

Law of Cosines:

If $△ABC$

has sides of length a , b , and c as shown, then:

  

  $\begin{array}{l}{a}^2=b^2+c^2-2bc\cos A\\ b^2=a^2+c^2-2ac\cos B\\ c^2=a^2+b^2-2ab\cos C\end{array}$

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:

The diagram representing given triangle is:

  

First find the angle opposite to the longest side, $\overline{AC}$ . Use the law of cosines to find angle B.

  $b^2=a^2+c^2-2ac\cos B$

Substitute the given values for a , b and c . And simplify,

  $\begin{array}{c}{16}^2={14}^2+9^2-2\left(14\right)\left(9\right)\cos B\\ \cos B=\frac{{16}^2-{14}^2-9^2}{-2\left(14\right)\left(9\right)}\\ \cos B\approx 0.0833\end{array}$

Use inverse cosine,

  $\begin{array}{l}B\approx {\cos }^{-1}\left(0.0833\right)\\ B\approx 85.22°\end{array}$

Now, by the law of Sines,

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin B}{b}\\ \frac{\sin A}{14}=\frac{\sin 85.22°}{16}\\ \sin A=\frac{14\sin 85.22°}{16}\\ \sin A\approx 0.872\end{array}$

Use inverse sine.

  $\begin{array}{l}A={\sin }^{-1}\left(0.875\right)\\ A\approx 60.69\end{array}$

Now, it is known that the sum of the three angles of a triangle is $180°$ . So,

  $\begin{array}{c}A+B+C=180°\\ 60.69°+85.22°+C\approx 180°\\ C\approx 180°-60.69°-85.22°\\ C\approx 34.09°\end{array}$