/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 43 A high fountain of water is loca... [FREE SOLUTION] | 91Ó°ÊÓ

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A high fountain of water is located at the center of a circular pool as shown in Figure P1.43. Not wishing to get his feet wet, a student walks around the pool and measures its circumference to be \(15.0 \mathrm{~m}\). Next, the student stands at the edge of the pool and uses a protractor to gauge the angle of elevation at the bottom of the fountain to be \(55.0^{\circ}\). How high is the fountain?

Short Answer

Expert verified
The height of the fountain is approximately 3.19 meters.

Step by step solution

01

Calculate the radius of the pool

The student measures the circumference of the pool to be \(15.0 \, m\). Given, Circumference C of a circle = \(2\pi r\). We can calculate radius r of the circle using the formula \(r = \frac{C}{2\pi}\). Substituting the given value of circumference, \(r = \frac{15.0}{2\pi} approx 2.39 \, m\). So, the radius of the pool is approximately 2.39 meters.
02

Calculate the height of the fountain

The student uses a protractor to measure the angle of elevation at the bottom of the fountain to be \(55^{\circ}\). The angle of elevation and adjacent side (radius) are known, and we are required to calculate the opposite side (height of fountain). This is a case of Tangent in trigonometry, where \(tan(\Theta) = \frac{opposite side}{adjacent side}\). So, the height (H) of the fountain can be calculated as \(H = r \cdot tan(\Theta)\). Substituting the known values, \(H = 2.39 \cdot tan(55) approx 3.19 \, m\). So, the height of the fountain is approximately 3.19 meters.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Circle Circumference
The circumference of a circle is a crucial concept in geometry, representing the distance around the circle. Easily understood as the circle's perimeter, the formula to calculate it is simple: \(C = 2\pi r\), where \(C\) is the circumference and \(r\) is the radius.

This formula shows how the circumference is directly proportional to the radius. Understanding this relationship is key: as the radius of the circle increases, so does its circumference.

In this problem, the student measured the circumference to be \(15\, \text{m}\). By rearranging the circumference formula to \(r = \frac{C}{2\pi}\), we can find the radius. Substituting \(C = 15\), we get \(r \approx 2.39\, \text{m}\). This gives us the radius without needing any other measurements.

Once you grasp this relationship, it becomes easier to tackle real-world problems involving circular shapes and sizes.
Tangent Function
The tangent function is a fundamental element of trigonometry. It relates the angles to the ratios between the sides in a right-angled triangle. Specifically, the tangent of an angle is the ratio of the opposite side to the adjacent side: \(\tan(\Theta) = \frac{\text{opposite side}}{\text{adjacent side}}\).

When a student measures the angle of elevation at the bottom of a fountain, they are dealing with a right-angled triangle where the radius of the pool is the adjacent side and the fountain height is the opposite side.

The student measured an angle of elevation at \(55^{\circ}\). With the radius known as approximately \(2.39\, \text{m}\), the height of the fountain can be found using the tangent function. The equation \(\tan(55^{\circ}) = \frac{\text{height}}{2.39}\) rearranges to find height: \(\text{height} = 2.39 \cdot \tan(55^{\circ})\).

This calculation results in the fountain's height being about \(3.19\, \text{m}\). Understanding tangent aids in solving problems involving right triangles, especially when involving heights and distances.
Angle of Elevation
The angle of elevation is the angle between the horizontal line (the line parallel to the ground) and the line of sight to an object above the horizontal line. It is always measured from the ground up.

In this scenario, a student measures the angle of elevation to the top of a fountain from the edge of a pool. This angle helps us determine how tall something is without measuring it physically.

Using the protractor, the student found an angle of elevation to be \(55^{\circ}\). This means that if you were standing on the ground looking up at the top of the fountain, your line of sight would create a \(55^{\circ}\) angle with your line running horizontal to the ground.

The angle of elevation combined with the radius (distance from the student to the base of the object) allows us to use trigonometric functions to calculate the height (opposite side) of the fountain. Calculating such angles in real-life situations can be quite handy for architects, engineers, and even hobbyists figuring out measurements from a distance.

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Most popular questions from this chapter

A right triangle has a hypotenuse of length \(3.00 \mathrm{~m}\), and one of its angles is \(30.0^{\circ}\). What are the lengths of (a) the side opposite the \(30.0^{\circ}\) angle and (b) the side adjacent to the \(30.0^{\circ}\) angle?

Assume that it takes \(7.00\) minutes to fill a \(30.0\)-gal gasoline tank. (a) Calculate the rate at which the tank is filled in gallons per second. (b) Calculate the rate at which the tank is filled in cubic meters per second. (c) Determine the time interval, in hours, required to fill a \(1.00-\mathrm{m}^{3}\) volume at the same rate. (1 U.S. gal \(=231\) in. \(^{3}\) )

(a) Find a conversion factor to convert from miles per hour to kilometers per hour. (b) For a while, federal law mandated that the maximum highway speed would be \(55 \mathrm{mi} / \mathrm{h}\). Use the conversion factor from part (a) to find the speed in kilometers per hour. (c) The maximum highway speed has been raised to \(65 \mathrm{mi} / \mathrm{h}\) in some places. In kilometers per hour, how much of an increase is this over the \(55-\mathrm{mi} / \mathrm{h}\) limit?

A woman measures the angle of elevation of a mountaintop as \(12.0^{\circ}\). After walking \(1.00 \mathrm{~km}\) closer to the mountain on level ground, she finds the angle to be \(14.0^{\circ}\). (a) Draw a picture of the problem, neglecting the height of the woman's eyes above the ground. Hint: Use two triangles. (b) Select variable names for the mountain height (suggestion: \(y\) ) and the woman's original distance from the mountain (suggestion: \(x\) ) and label the picture. (c) Using the labeled picture and the tangent function, write two trigonometric equations relating the two selected variables. (d) Find the height \(y\) of the mountain by first solving one equation for \(x\) and substituting the result into the other equation.

Assume that there are 100 million passenger cars in the United States and that the average fuel consumption is \(20 \mathrm{mi} / \mathrm{gal}\) of gasoline. If the average distance traveled by each car is \(10000 \mathrm{mi} / \mathrm{yr}\), how much gasoline would be saved per year if average fuel consumption could be increased to \(25 \mathrm{mi} / \mathrm{gal}\) ?

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