Chapter 9.6, Problem 6E

Solution

To decide: that to solve the triangle with the given case whether the law of sines or the law of cosines should be used.

Law of cosines

Given information:

The given case is:

SAS

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:

To use the law of sines measure of lengths of two sides and an angle opposite one of the two sides must be known. Since here the measure is known of two sides of the triangle but the measure of angle that is known is the angle included by these two sides and not that of the opposite angle. So, here law of sines can’t be used.

Law of cosines to solve this triangle as the measure of two sides and the included angle is known. As first law of cosines can be used to find the measure of the third side, and then once the measure of all three sides are known, the law of cosines can be used to find the measure of rest two angles.

Therefore, in this case, the law of cosines should be used.