Chapter 4.3, Problem 76E

a.

To determine

Use a variable to indicate the height of the balloon.

Answer to Problem 76E

  $\boxed{x}$

Explanation of Solution

Given information:

A $20$ -meter line is a tether for a helium-filled balloon. Because of a breeze, the line makes an angle of approximately $85°$ with the ground.

Show the known quantities of the triangle and use a variable to indicate the height of the balloon.

Calculation:

The height of the balloon is represented by $x$ .

b.

To determine

Solve an equation for the height of the balloon.

Answer to Problem 76E

  $\boxed{\sin 85°=\frac{x}{20}}$

Explanation of Solution

Given information:

A $20$ -meter line is a tether for a helium-filled balloon. Because of a breeze, the line makes an angle of approximately $85°$ with the ground.

Use a trigonometric function to write and solve an equation for the height of the balloon.

Calculation:

we can see that the length of the hypotenuse of the triangle is $20$ meter.

The length of the side opposite to the angle $85°$ is given by $x$ .

Now, we know that a trignometric function which involves the two sides, that is the hypotenuse and the opposite side to the angle is the sine function.

Consider a right triangle, with one acute angle $\theta$ , then

  $\sin \theta =\frac{\text{Opposite-side}}{\text{Hypotenuse}}$

Substituting the values in the above given formula of sin, we have,

  $\sin 85°=\frac{x}{20}$

Hence, the equation invoolving the unknown quantity $x$ is $\boxed{\sin 85°=\frac{x}{20}}$ .

c.

To determine

How does this affect the triangle you drew in part (a)?

Answer to Problem 76E

The angle the ballon makes with the ground becomes smaller.

Explanation of Solution

Given information:

A $20$ -meter line is a tether for a helium-filled balloon. Because of a breeze, the line makes an angle of approximately $85°$ with the ground.

The breeze becomes stronger and the angle the line makes with the ground decreases. How does this affect the triangle you drew in part (a)?

Calculation:

In order to find the height of the balloon, $x$ , we have to solve the equation, $\sin 85°=\frac{x}{20}$ .

Hence, $x=20\times \sin 85°$ .

Now, $\sin 85°=0.9962$ .

Substitute this in equation of $x$ , we get,

  $x=20\times \sin 85°$

  $\begin{array}{l}=20\times 0.9962\\ =19.92\end{array}$

Hence, the height of the ballon is $\boxed{x=19.92}$ .

As the breeza becomes stronger, the hypotenuse of the triangle drwan above tilts more to the ground and hence the angle the ballon makes with the ground becomes smaller. As such the height of the ballon also decreases.

angle $\theta <85°$

Hence, the angle the ballon makes with the ground becomes smaller.

d.

To determine

Complete the table, which shows the heights (in meters) of the balloon for decreasing angle measures $\theta$ .

Answer to Problem 76E

  $\boxed{19.696meters}$ , $\boxed{18.794meters}$ , $\boxed{17.321meters}$ , $\boxed{15.321meters}$ , $\boxed{12.856meters}$ , $\boxed{10meters}$ , $\boxed{6.84meters}$ and $\boxed{3.473meters}$

Explanation of Solution

Given information:

A $20$ -meter line is a tether for a helium-filled balloon. Because of a breeze, the line makes an angle of approximately $85°$ with the ground.

Complete the table, which shows the heights (in meters) of the balloon for decreasing angle measures $\theta$ .

Calculation:

We have to find the height of the ballon for decreasing angle measures $\theta$

  

For

  $\theta =80°$

Let us use the equation involving the height $x$ of the ballon, which is,

  $\begin{array}{l}x=20\times \sin 80°\\ =20\times 0.9848\\ =19.696\end{array}$

Hence, for $\theta =80°$ , height of the balloon is $\boxed{19.696meters}$ .

For

  $\theta =70°$

Let us use the equation involving the height $x$ of the ballon, which is,

  $\begin{array}{l}x=20\times \sin 70°\\ =20\times 0.9396\\ =18.794\end{array}$

Hence, for $\theta =70°$ , height of the balloon is $\boxed{18.794meters}$

For

  $\theta =60°$

Let us use the equation involving the height $x$ of the ballon, which is,

  $\begin{array}{l}x=20\times \sin 60°\\ =20\times 0.866\\ =17.321\end{array}$

Hence, for $\theta =60°$ , height of the balloon is $\boxed{17.321meters}$

For

  $\theta =50°$

Let us use the equation involving the height $x$ of the ballon, which is,

  $\begin{array}{l}x=20\times \sin 50°\\ =20\times 0.766\\ =15.321\end{array}$

Hence, for $\theta =50°$ , height of the balloon is $\boxed{15.321meters}$

For

  $\theta =40°$

Let us use the equation involving the height $x$ of the ballon, which is,

  $\begin{array}{l}x=20\times \sin 40°\\ =20\times 0.643\\ =12.856\end{array}$

Hence, for $\theta =40°$ , height of the balloon is $\boxed{12.856meters}$

For

  $\theta =30°$

Let us use the equation involving the height $x$ of the ballon, which is,

  $\begin{array}{l}x=20\times \sin 30°\\ =20\times 0.5\\ =10\end{array}$

Hence,for $\theta =30°$ , height of the balloon is $\boxed{10meters}$

For

  $\theta =20°$

Let us use the equation involving the height $x$ of the ballon, which is,

  $\begin{array}{l}x=20\times \sin 20°\\ =20\times 0.342\\ =6.84\end{array}$

Hence, for $\theta =20°$ , height of the ballon is $\boxed{6.84meters}$

For

  $\theta =10°$

Let us use the equation involving the height $x$ of the ballon, which is,

  $\begin{array}{l}x=20\times \sin 10°\\ =20\times 0.1736\\ =3.473\end{array}$

Hence, for $\theta =10°$ , height of the balloon is $\boxed{3.473meters}$

e.

To determine

As $\theta$ approaches $0°$ , how does this affect the height of the balloon?

Answer to Problem 76E

The height also approaches zero.

Explanation of Solution

Given information:

A $20$ -meter line is a tether for a helium-filled balloon. Because of a breeze, the line makes an angle of approximately $85°$ with the ground.

As $\theta$ approaches $0°$ , how does this affect the height of the balloon?

Calculation:

This is because, the height $x$ of the balloon is given by,

  $\begin{array}{l}x=20\times \sin 0°\\ =20\times 0\\ =0\end{array}$

Hence, we can clearly see that as $\theta \to 0$ , the height also approaches zero.