Chapter 4.3, Problem 84E

(a).

To determine

To draw:

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

Explanation of Solution

Given:

A $20$ meter line is used to tether a helium filled balloon. Because of a breeze, the line makes an angle of approximately ${85}^{\circ }$ with the ground.

Draw:

Use the given formula and draw a right triangle where hypotenuse is $20$ meter line and the angle made by the hypotenuse with the ground is ${85}^{\circ }$ .

  

(b).

To determine

To find:

An equation involving the unknown quantity.

Answer to Problem 84E

The required equation involving the unknown quantity is $\sin {85}^{\circ }=\frac{h}{20}$ .

Explanation of Solution

Given:

A $20$ meter line is used to tether a helium filled balloon. Because of a breeze, the line makes an angle of approximately ${85}^{\circ }$ with the ground.

Calculation:

  

Apply the formula for the sine trigonometric ratio:

  $\sin {85}^{\circ }=\frac{P}{H}$

Substitute $P=h$ and $H=20$ in the above formula:

  $\sin {85}^{\circ }=\frac{h}{20}$

Conclusion:

Hence, the required equation involving the unknown quantity is $\sin {85}^{\circ }=\frac{h}{20}$ .

(c).

To determine

To find:

The height of the balloon.

Answer to Problem 84E

The height of the balloon is $19.92$ meters.

Explanation of Solution

Given:

A $20$ meter line is used to tether a helium filled balloon. Because of a breeze, the line makes an angle of approximately ${85}^{\circ }$ with the ground.

  

From the solution of part(b):

  $\sin {85}^{\circ }=\frac{h}{20}$

Solve the above equation for $h$ :

  $\begin{array}{c}h=20\left(\sin {85}^{\circ }\right)\\ \approx 20\left(0.9961\right)\\ \approx 19.92\end{array}$

Hence, the height of the balloon is $19.92$ meters.

(d).

To determine

To explain:

If the angle the balloon makes with the ground decreases does this affect the triangle or not which is drew in part(a).

Answer to Problem 84E

The height of the triangle decreases and the base of the triangle increases.

Explanation of Solution

Given:

A $20$ meter line is used to tether a helium filled balloon. Because of a breeze, the line makes an angle of approximately ${85}^{\circ }$ with the ground.

  

If the angle the balloon makes with the ground decreases then the height of the triangle decreases and the base of the triangle increases.

Hence, the height of the triangle decreases.

(e).

To determine

To find:

The heights of the balloon for decreasing angle measures $\theta$ .

Answer to Problem 84E

The required table is:

Angle,   $\theta$	${80}^{\circ }$	${70}^{\circ }$	${60}^{\circ }$	${50}^{\circ }$	${40}^{\circ }$	${30}^{\circ }$	${20}^{\circ }$	${10}^{\circ }$
Height	$19.696$	$18.792$	$17.32$	$15.32$	$12.854$	$10$	$6.84$	$3.472$

Explanation of Solution

Given:

A $20$ meter line is used to tether a helium filled balloon. Because of a breeze, the line makes an angle of approximately ${85}^{\circ }$ with the ground.

  

The height of the balloon is:

  $h=20\left(\sin \theta \right)$

Assume $\theta ={80}^{\circ }$ then height of the balloon is:

  $\begin{array}{c}h=20\left(\sin {80}^{\circ }\right)\\ =20\left(0.9848\right)\\ \approx 19.696\end{array}$

Assume $\theta ={70}^{\circ }$ then height of the balloon is:

  $\begin{array}{c}h=20\left(\sin {70}^{\circ }\right)\\ =20\left(0.9396\right)\\ \approx 18.792\end{array}$

Assume $\theta ={60}^{\circ }$ then height of the balloon is:

  $\begin{array}{c}h=20\left(\sin {60}^{\circ }\right)\\ =20\left(0.8660\right)\\ \approx 17.32\end{array}$

Assume $\theta ={50}^{\circ }$ then height of the balloon is:

  $\begin{array}{c}h=20\left(\sin {50}^{\circ }\right)\\ =20\left(0.7660\right)\\ \approx 15.32\end{array}$

Assume $\theta ={40}^{\circ }$ then height of the balloon is:

  $\begin{array}{c}h=20\left(\sin {40}^{\circ }\right)\\ =20\left(0.6427\right)\\ \approx 12.854\end{array}$

Assume $\theta ={30}^{\circ }$ then height of the balloon is:

  $\begin{array}{c}h=20\left(\sin {30}^{\circ }\right)\\ =20\left(0.5\right)\\ \approx 10\end{array}$

Assume $\theta ={20}^{\circ }$ then height of the balloon is:

  $\begin{array}{c}h=20\left(\sin {20}^{\circ }\right)\\ =20\left(0.3420\right)\\ \approx 6.84\end{array}$

Assume $\theta ={10}^{\circ }$ then height of the balloon is:

  $\begin{array}{c}h=20\left(\sin {10}^{\circ }\right)\\ =20\left(0.1736\right)\\ \approx 3.472\end{array}$

Hence, the required table is:

Angle,   $\theta$	${80}^{\circ }$	${70}^{\circ }$	${60}^{\circ }$	${50}^{\circ }$	${40}^{\circ }$	${30}^{\circ }$	${20}^{\circ }$	${10}^{\circ }$
Height	$19.696$	$18.792$	$17.32$	$15.32$	$12.854$	$10$	$6.84$	$3.472$