Chapter 6.2, Problem 11E

To determine

To find: The solution for the triangle by law of cosine.

Answer to Problem 11E

The angles of the triangle are $A=31.40°$ , $B=128.60°$ and $C=6.56°$ .

Explanation of Solution

Given:

The given diagram is shown in Figure 1

  

Figure 1

Calculation:

Consider the sides of the triangle is,

  $\begin{array}{l}a=8\text{mm}\\ b=18\text{mm}\\ c=12\text{mm}\end{array}$

Consider the standard form for the law of cosine is,

  $b^2=a^2+c^2-2ac\cos B$

Then,

  $\begin{array}{c}{b}^2={\left(10\right)}^2+{\left(15\right)}^2-2\left(15\right)\left(10\right)\cos \left(20°\right)\\ =6.56\text{mm}\end{array}$

Consider the reciprocal form of law of sine is,

  $\frac{\sin A}{a}=\frac{\sin B}{b}$

Then,

  $\begin{array}{l}\frac{\sin A}{10}=\frac{\sin 20°}{6.56}\\ \sin A=0.5214\\ A=31.40°\end{array}$

There are two angles between $0°$ and $180°$ as,

  $\begin{array}{c}{A}_2=180°-31.43°\\ =148.57°\end{array}$

Consider the value of angle $C_1$ for $A=31.40°$ by angle sum property,

  $\begin{array}{c}{C}_1=180°-31.40°-20°\\ =128.60°\end{array}$

Consider the value of angle $C_2$ for $A_2=148.57°$ by angle sum property,

  $\begin{array}{c}{C}_2=180°-148.57°-20°\\ =11.43°\end{array}$