Chapter 6.1, Problem 51E

(a)

To determine

To Find: The diagram to represent the given data.

Answer to Problem 51E

The required diagram is shown in Figure 1

Explanation of Solution

Given:

The angle of elevation to an airplane from two points A and B on level ground are $55°$ and $72°$ respectively. The points A and B are 2.2 miles apart and the airplane is the east of both the points is the same vertical plane.

Calculation:

From the given data the required diagram is shown in Figure 1

  

Figure 1

(b)

To determine

To Find: The distance between the plane and the point B .

Answer to Problem 51E

The distance between point B and the plane is $6.16\text{km}$ .

Explanation of Solution

Consider the angle $\angle ABD$ is obtained as,

  $\begin{array}{c}\angle ABD=77°-55°\\ =17°\end{array}$

Obtain the distance BD by the law of sine as,

  $\begin{array}{l}\frac{BD}{\sin 55°}=\frac{AB}{\sin 17°}\\ \frac{BD}{\sin 55°}=\frac{2.2}{\sin 17°}\\ BD=\frac{2.2\sin 55°}{\sin 17°}\\ BD=6.16\text{km}\end{array}$

Thus, the distance between point B and the plane is $6.16\text{km}$ .

(c)

To determine

To Find: The altitude of the plane.

Answer to Problem 51E

The altitude of the plane is $6.16\text{miles}$ .

Explanation of Solution

Consider the value of angle C is obtained as,

  $\begin{array}{c}C=180°-\left[A+B\right]\\ =180°-\left[50°+\left(180°-72°\right)\right]\\ =17°\end{array}$

Then, the distance BC by sine angle property is obtained as,

  $\begin{array}{l}\frac{2.2}{\sin 17°}=\frac{BC}{\sin 55°}\\ BC=\frac{2.2\sin 55°}{\sin 17°}\\ BC=6.16\text{miles}\end{array}$

(d)

To determine

To Find: The distance the plane must travel before it is directly above the point A.

Answer to Problem 51E

The distance that the plane must travel is $4.10\text{miles}$ .

Explanation of Solution

Consider the distance from AD is obtained as,

  $\begin{array}{c}AD=BD+AB\\ =2.2+BD\end{array}$

Then, use the cosine angle as,

  $\begin{array}{l}\cos 72°=\frac{BD}{BC}\\ BD=\frac{BD}{6.16°}\\ BD=6.16\cos 72°\end{array}$

Then,

  $\begin{array}{c}AD=2.2+6.16\cos 72°\\ =4.10\text{miles}\end{array}$