Chapter 6.2, Problem 12E

To determine

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

Answer to Problem 12E

The angles of the triangle are $A=53°30^{\prime }$ , $B=50°30^{\prime }$ and $C=3°15^{\prime }$ .

Explanation of Solution

Given:

The given diagram is shown in Figure 1

  

Figure 1

The sides of the triangle is,

  $\begin{array}{l}a=10\text{ft}\\ c=12.5\text{ft}\end{array}$

The angels of the triangles is,

  $B=50°30^{\prime }$

Calculation:

Consider the standard form for the law of cosine is,

  $b^2=a^2+b^2-2ac\cos B$

Then,

  $\begin{array}{c}{b}^2={\left(10.4\right)}^2+{\left(12.5\right)}^2-2\left(10.4\right)\left(12.5\right)\cos \left(50°30^{\prime }\right)\\ =9.95\text{ft}\end{array}$

Consider the reciprocal form of law of sine is,

  $\frac{\sin A}{a}=\frac{\sin B}{b}$

Then,

  $\begin{array}{l}\frac{\sin A}{10.4}=\frac{\sin 50°30^{\prime }}{9.95}\\ \sin A=0.807\\ A=53°45^{\prime }\end{array}$

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

  $\begin{array}{c}{A}_2=180°-53°45^{\prime }\\ =126°15^{\prime }\end{array}$

Consider the value of angle $C_1$ for $A=53°45^{\prime }$ by angle sum property,

  $\begin{array}{c}{C}_1=180°-50°30^{\prime }-53°45^{\prime }\\ =75°45^{\prime }\end{array}$

Consider the value of angle $C_2$ for $A_2=126°15^{\prime }$ by angle sum property,

  $\begin{array}{c}{C}_2=180°-50°30^{\prime }-126°15^{\prime }\\ =3°15^{\prime }\end{array}$