Chapter 9.6, Problem 17E

Solution

To solve: the $△ABC$ .

  $A\approx 42.6°,B\approx 11.7°,C\approx 125.7°$

Given information:

The given measurements are:

  $a=10,b=3,c=12$

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:

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,

  $\begin{array}{c}{3}^2={10}^2+{12}^2-2\left(10\right)\left(12\right)\cos B\\ \cos B=\frac{3^2-{10}^2-{12}^2}{-2\left(10\right)\left(12\right)}\\ \cos B\approx 0.9792\end{array}$

Use inverse cosine,

  $\begin{array}{l}B\approx {\cos }^{-1}\left(0.9792\right)\\ B\approx 11.7°\end{array}$

Now, by the law of Sines,

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin B}{b}\\ \frac{\sin A}{10}=\frac{\sin 11.7°}{3}\\ \sin A=\frac{10\sin 11.7°}{3}\\ \sin A\approx 0.6759\end{array}$

Use inverse sine.

  $\begin{array}{l}A={\sin }^{-1}\left(0.6759\right)\\ A\approx 42.6°\end{array}$

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

  $\begin{array}{c}A+B+C=180°\\ 42.6°+11.7°+C\approx 180°\\ C\approx 180°-42.6°-11.7°\\ C\approx 125.7°\end{array}$