Chapter 6.1, Problem 45E

(a)

To determine

Todraw:The diagram that illustrate the problem.

Explanation of Solution

The diagram that illustrate the problem is shown below.

  

Figure (1)

Therefore, the required diagram is shown in Figure (1).

(b)

To determine

To write:The equation for $d$ .

Explanation of Solution

The law of sine is given by,

  $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

Consider figure (1).

  $\begin{array}{l}A=\theta \\ B=180°-\varphi \end{array}$

The value of $C$ is calculated as,

  $\begin{array}{c}C=180°-A-B\\ =180°-\theta -180°+\varphi \\ =\varphi -\theta \end{array}$

Now by the law of sine, the distance between the airplane and the point $B$ is calculated as,

  $\begin{array}{c}\frac{2}{\sin \left(\varphi -\theta \right)}=\frac{d}{\sin \theta }\\ d=\frac{2\sin \theta }{\sin \left(\varphi -\theta \right)}\end{array}$

Therefore, the expression for $d$ is $d=\frac{2\sin \theta }{\sin \left(\varphi -\theta \right)}$ .

(c)

To determine

To find:The distance from point B.

Explanation of Solution

The expression for $d$ is given by,

  $d=\frac{2\sin \theta }{\sin \left(\varphi -\theta \right)}$

Substitute the values in the above expression.

  $\begin{array}{c}d=\frac{2\sin 40°}{\sin \left(60°-40°\right)}\\ =3.76\end{array}$

Therefore, the distance of the plane from point B is $3.76\text{miles}$ .