Chapter 4, Problem 5PS

a.

To determine

Find the shortest distance $d$ the helicopter must travel to land on the island.

Answer to Problem 5PS

  $\boxed{4767feet}$

Explanation of Solution

Given information:

A surveyor in a helicopter is determining the width of an island, as shown in the figure.

  

What is the shortest distance $d$ the helicopter must travel to land on the island?

Calculation:

Here, we will consider the following diagram:

  

Here, $W$ is the width of the island.

Now, we will apply trigonometric identities in the smaller right angled triangle:

  $\sin 39°=\frac{3000}{d}$

Thus,

  $\begin{array}{l}d=\frac{3000}{\sin 39°}\\ d=\frac{3000}{0.63}\\ d=\boxed{4767feet}\end{array}$

Hence, the shortest distance is $\boxed{4767feet}$ .

b.

To determine

Find the horizontal distance $x$ the helicopter must travel before it reaches the nearer end of the island.

Answer to Problem 5PS

  $\boxed{3705feet}$

Explanation of Solution

Given information:

A surveyor in a helicopter is determining the width of an island, as shown in the figure.

  

What is the horizontal distance $x$ the helicopter must travel before it is directly over the nearer end of the island?Calculation:

Here, we will apply trigonometric identities in the smaller triangle to obtain $x$ :

  $\tan 39°=\frac{3000}{x}$

Thus,

  $\begin{array}{l}x=\frac{3000}{\tan 39°}\\ x=\frac{3000}{0.81}\\ x=\boxed{3705feet}\end{array}$

Hence, the horizontal distance is $\boxed{3705feet}$ .

c.

To determine

Find the width $W$ of the island.

Answer to Problem 5PS

  $\boxed{2183feet}$

Explanation of Solution

Given information:

A surveyor in a helicopter is determining the width of an island, as shown in the figure.

  

Find the width $W$ of the island. Explain how you found your answer.Calculation:

Here, we will apply trigonometric identities in the bigger triangle to obtain $W$ :

  $\begin{array}{l}\tan 27°=\frac{3000}{x+W}\\ \tan 27°=\frac{3000}{3705+W}\end{array}$

Thus,

  $\begin{array}{l}3705+W=\frac{3000}{\tan 27°}\\ 3705+W=\frac{3000}{0.51}\\ 3705+W=5882.35\\ W=\boxed{2183feet}\end{array}$

Hence, the width $W$ of the island is $\boxed{2183feet}$ .