Chapter 6.4, Problem 42E

View from a Satellite The figures on the next page indicate that the higher the orbit of a satellite, the more of the earth the satellite can “see.” Let θ, s, and h be as in the figure, and assume that the earth is a sphere of radius 3960 mi.

1. (a) Express the angle θ as a function of h.
2. (b) Express the distance s as a function of θ.
3. (c) Express the distance s as a function of h. [Hint: Find the composition of the functions in parts (a) and (b).]
4. (d) If the satellite is 100 mi above the earth, what is the distance s that it can see?
5. (e) How high does the satellite have to be to see both Los Angeles and New York, 2450 mi apart?

To determine

To Express: The angle of elevation $\theta$ as a function of $h$.

Answer to Problem 42E

The expression that represents the angle of elevation $\theta$ of the sun as a function of length s of the shadow of pole is $\theta ={\tan }^{-1}\left(\frac{50}{s}\right)$.

Explanation of Solution

Given:

The below figure show the length $50\text{ft}$ casts a shadow of length s that makes an angle of elevation $\theta$.

Formula Used: Trigonometric Property of tangent:

$\tan \theta =\frac{\text{opposite}}{\text{adjecant}}$

Calculation:

Trigonometric Property of tangent is,

$\tan \theta =\frac{\text{opposite}}{\text{adjacent}}$

Substitute s for adjacent and $50$ for opposite in the above formula,

$\tan \theta =\frac{50}{s}$

Simplify further to get the value of $\theta$ by taking ${\tan }^{-1}$ both sides in the above expression.

$\begin{array}{c}{\tan }^{-1}\left(\tan \theta \right)={\tan }^{-1}\left(\frac{50}{s}\right)\\ \theta ={\tan }^{-1}\left(\frac{50}{s}\right)\left(\because {\tan }^{-1}\left(\tan \theta \right)=\theta \right)\end{array}$

Thus, the expression that represents the angle of elevation $\theta$ of the sun as a function of length s of the shadow of pole is $\theta ={\tan }^{-1}\left(\frac{50}{s}\right)$.

To determine

To Find: The angle of elevation $\theta$ of sun.

Answer to Problem 42E

The angle of elevation of the sun when the shadow of the pole is $20\text{ft}$ is $1.19$.

Explanation of Solution

Given:

The length of the shadow is $s=20\text{ft}$ and the length of the pole is $50\text{ft}$.

Calculation:

From part (a) we get the expression,

$\theta ={\tan }^{-1}\left(\frac{50}{s}\right)$

Substitute $20$ for s in the above expression,

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{50}{20}\right)\\ ={\tan }^{-1}\left(\frac{5}{2}\right)\\ =68.1°\left(\because {\tan }^{-1}\left(\frac{5}{2}\right)=68.1°\right)\end{array}$

Thus, the angle of elevation of the sun when the shadow of the pole is $20\text{ft}$ is $68.1°$.