Chapter 6.2, Problem 18E

To determine

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

Answer to Problem 18E

The angles of the triangle are $B=10.53°$ and $C=121.47°$ , the side $a$ of the triangle is $12.19$ .

Explanation of Solution

Given:

The sides of the triangle is,

  $\begin{array}{l}b=3\\ c=14\end{array}$

The angle $A$ of the triangle is $A=48°$ .

Calculation:

Consider the standard form for the law of cosine is,

  $a^2=b^2+c^2-2bc\cos A$

Then,

  $\begin{array}{l}{a}^2={\left(3\right)}^2+{\left(14\right)}^2-2\left(3\right)\left(14\right)\cos 48°\\ a^2=205-84\cos 48°\\ a=12.19\end{array}$

Consider the reciprocal form of law of sine is,

  $\frac{\sin A}{a}=\frac{\sin B}{b}$

Then,

  $\begin{array}{l}\frac{\sin B}{6}=\frac{\sin 48°}{12.19}\\ \sin B=0.1829\\ B=10.53°\end{array}$

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

  $\begin{array}{c}{B}_2=180°-10.53°\\ =169.47°\end{array}$

Consider the value of angle $C_1$ for $B=10.53°$ by angle sum property,

  $\begin{array}{c}{C}_1=180°-48°-10.53°\\ =121.47°\end{array}$

Consider the value of angle $C_2$ for $B_2=169.47°$ by angle sum property,

  $\begin{array}{c}{C}_2=180°-48°-173.11°\\ =-41.11°\end{array}$