Chapter 4.3, Problem 59E

Solution

a.

To identify: The distance d from the group to the jet for the given angle. $\theta =45°$

An F-15 aircraft flying at an altitude of 8000ft passes directly over a group of vacationers hiking at 7400 ft. If ? is the angle of elevation from the hikers to the F-15,

The displacement of the weight at $t=0$ is $0.4inches$

Given information:

The given is jet is flying at the altitude of 8000ft and vacationers are hiking at 7400ft

Angle of evevation from the hikers to the F − 15 is $\theta$

  

Explanation:

Consider the given information.

Consider the triangle ABC

  $\begin{array}{l}\sin \theta =\frac{\text{opposite}\text{side}}{\text{hypotenuse}}\\ \sin 45°=\frac{600}{d}\\ d=\frac{600}{\sin 45°}\\ d=\frac{600}{\left(\frac{1}{\sqrt{2}}\right)}\\ d\approx 848.53\end{array}$

Hence the distance d from the group to the aircraft at $\theta =45°$ is $848.53\text{ft}\text{.}$

b.

To identify: The distance d from the group to the jet for the given angle. $\theta =90°$

The distance d from the group to the aircraft at $\theta =90°$ is $\text{600ft}\text{.}$ .

Given information:

The given is jet is flying at the altitude of 8000ft and vacationers are hiking at 7400ft

Angle of evevation from the hikers to the F − 15 is $\theta$ .

  

Explanation:

Consider the given information.

Consider the triangle ABC

  $\begin{array}{c}\sin \theta =\frac{\text{opposite}\text{side}}{\text{hypotenuse}}\\ \sin 90°=\frac{600}{d}\\ d=\frac{600}{\sin 90°}\\ d=\frac{600}{1}\\ d=600\end{array}$

Hence the distance d from the group to the aircraft at $\theta =90°$ is $\text{600ft}\text{.}$ .

c.

To identify: The distance d from the group to the jet for the given angle $\theta =140°$ .

The distance d from the group to the aircraft at $\theta =140°$ is $\text{933}\text{.43ft}\text{.}$

Given information:

The given is jet is flying at the altitude of 8000ft and vacationers are hiking at 7400ft.

Angle of evevation from the hikers to the F − 15 is $\theta$ .

  

Explanation:

Consider the given information.

Consider the triangle ABC

  $\begin{array}{c}\sin \theta =\frac{\text{opposite}\text{side}}{\text{hypotenuse}}\\ \sin 140°=\frac{600}{d}\\ d=\frac{600}{\sin 140°}\\ d\approx 933.43\end{array}$

Hence the distance d from the group to the aircraft at $\theta =140°$ is $\text{933}\text{.43ft}$ .