Chapter 4.3, Problem 72E

a

To determine

Draw a right triangle that gives a visual representation of the problem.

Answer to Problem 72E

  $\boxed{height=X}$

Explanation of Solution

Given information:

A $20$ -meter line is used to tether a helium-filled balloon. The line makes an angle of approximately 85° with the ground because of a breeze.

Draw a right triangle that gives a visual representation of the problem. Label the known quantities of the triangle and use a variable to represent the height of the balloon.

Calculation:

A $20$ -meter line is used to tether a helium-filled balloon. Due to breeze the line makes an angle of approximately $85°$ with the ground. We draw a diagram below,

The height of the balloon is $X$ .

  

Hence, the height of balloon is $X$ .

b.

To determine

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

Answer to Problem 72E

  $\boxed{\sin 85°=\frac{x}{20}}$

Explanation of Solution

Given information:

A $20$ -meter line is used to tether a helium-filled balloon. The line makes an angle of approximately 85° with the ground because of a breeze.

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

Calculation:

A $20$ -meter line is used to tether a helium-filled balloon. Due to breeze the line makes an angle of approximately $85°$ with the ground. We draw a diagram below,

The height of the balloon is $X$ .

  

From the figure length of hypotenuse of the triangle is $20$ meters.

Consider e right triangle with one acute angle $\theta$ so,

  $\begin{array}{l}\sin \theta =\frac{opposite,side}{hypotenuse}\\ \sin 85°=\frac{x}{20}\end{array}$

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

c.

To determine

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)?

Answer to Problem 72E

  $\boxed{x=19.92}$ meters.

Explanation of Solution

Given information:

A $20$ -meter line is used to tether a helium-filled balloon. The line makes an angle of approximately 85° with the ground because of a breeze.

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:

A $20$ -meter line is used to tether a helium-filled balloon. Due to breeze the line makes an angle of approximately $85°$ with the ground. We draw a diagram below,

The height of the balloon is $X$ .

  

From the figure length of hypotenuse of the triangle is $20$ meters.

In order to find the height of the balloon,

  $\begin{array}{l}\sin 85°=\frac{x}{20}\\ x=20\times \sin 85°\\ x=20\times 0.9962\\ x=19.92\end{array}$

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

d.

To determine

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

Answer to Problem 72E

  $\begin{array}{ccccccccc}Angle,\theta & 80°& 70°& 60°& 50°& 40°& 30°& 20°& 10°\\ Height& 19.696& 18.794& 17.321& 15.321& 12.856& 10& 6.84& 3.43\end{array}$

Explanation of Solution

Given information:

A $20$ -meter line is used to tether a helium-filled balloon. The line makes an angle of approximately 85° with the ground because of a breeze.

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

  $\begin{array}{ccccccccc}Angle,\theta & 80°& 70°& 60°& 50°& 40°& 30°& 20°& 10°\\ Height& & & & & & & & \end{array}$

Calculation:

Consider the equation involving the height $X$ of the balloon,

  $\theta =80°$

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

Now $\theta =70°$

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

Now, $\theta =60°$

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

Now,

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

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

Now,

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

Now,

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

Now,

  $\begin{array}{l}\theta =10°\\ x=20\times \sin 10°\\ x=20\times 0.1736\\ \boxed{x=3.43}meters\end{array}$

Hence the table is $\begin{array}{ccccccccc}Angle,\theta & 80°& 70°& 60°& 50°& 40°& 30°& 20°& 10°\\ Height& 19.696& 18.794& 17.321& 15.321& 12.856& 10& 6.84& 3.43\end{array}$ .

e.

To determine

As $\theta$ approaches $0°$ , how does this affect the height of the balloon? Draw a right triangle to explain your reasoning.

Answer to Problem 72E

  $\boxed{x=0}meters$

Explanation of Solution

Given information:

A $20$ -meter line is used to tether a helium-filled balloon. The line makes an angle of approximately 85° with the ground because of a breeze.

As $\theta$ approaches $0°$ , how does this affect the height of the balloon? Draw a right triangle to explain your reasoning.

Calculation:

As $\theta$ approaches $0°$ , the height of the balloon is,

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

Hence, the height of the balloon is $\boxed{x=0}meters$ .