Chapter 8, Problem 25RE

To determine

To find: The remaining angle(s) and side(s) of the triangle, if it(they) exist.

Answer to Problem 25RE

The remaining side is $c=2.33$ and the angles are $A=16.08°$ and $B=123.92°$ .

Explanation of Solution

Given:

The sides $a=1$ , $b=3$ and the angle $C=40°$ .

Calculation:

The Cosine Law for a triangle $\Delta ABC$ is given by the following equations:

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

Here, the capital letters represent the angles and the small letters represent the length of the respective sides of the triangle.

Given the 2 sides a and b , and the angle C , the third side c can be determined as follows:

  $\begin{array}{c}{c}^2=a^2+b^2-2ab\left(\cos C\right)\\ =1+9-6\left(\cos 40°\right)\\ =5.404\\ c=2.33\end{array}$

With the 3 sides known, the remaining 2 angles can be determined as follows:

  $\begin{array}{c}{a}^2=b^2+c^2-2bc\left(\cos A\right)\\ 1=9+5.404-4.66\left(\cos A\right)\\ \cos A=\frac{13.404}{4.66}\\ A=16.08°\end{array}$

  $\begin{array}{c}{b}^2=a^2+c^2-2ac\left(\cos B\right)\\ 9=1+5.404-4.66\left(\cos B\right)\\ \cos B=\frac{3.596}{4.66}\\ B=123.92°\end{array}$

So, the remaining side is $c=2.33$ and the angles are $A=16.08°$ and $B=123.92°$ .