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Chapter 1: Problem 68

friend lives in the other building. The two of you are having a discussion about the heights of the buildings, and your friend claims that the height of his building is more than 1.50 times the height of yours. To resolve the issue you climb to the roof of your building and estimate that your line of sight to the top edge of the other building makes an angle of \(21^{\circ}\) above the horizontal, whereas your line of sight to the base of the other building makes an angle of \(52^{\circ}\) below the horizontal. Determine the ratio of the height of the taller building to the height of the shorter building. State whether your friend is right or wrong.

Short Answer

Expert verified
The ratio is approximately 1.3; your friend is incorrect.

Step by step solution

01

Understand the Problem

We need to find the ratio of the height of the taller building (your friend's building) to the height of the shorter one (your building) using angles of elevation and depression. These angles help us establish the respective heights of the buildings based on your position.
02

Analyze the Angles

You observe the top of your friend's building at an angle of elevation of \(21^{\circ}\) and the base at an angle of depression of \(52^{\circ}\). This means the total angular displacement between your line of sight to the top and the base of the other building is \(21^{\circ} + 52^{\circ} = 73^{\circ}\).
03

Use Trigonometry to Find Building Heights

We can express the height of the other building, \(H\), and the height of your building, \(h\), using trigonometric ratios. Considering your observations involve you looking upwards and downwards from your building, we can use tangent functions for both angles to relate the heights and the distance between the buildings.
04

Establish the Height Equations

Let's consider the distance between the two buildings as \(d\). The height of your friend's building from the base to its top \((H)\) can be expressed as: \[H = d \tan(21^{\circ}) + d \tan(52^{\circ}) = d (\tan(21^{\circ}) + \tan(52^{\circ})).\]
05

Height of Your Building

The height of your building \(h\) can be considered equal to the base height from which both angles are measured. Thus, given the angle of depression to the base: \[h = d \tan(52^{\circ}).\]
06

Calculate the Ratio of Heights

Now, form the ratio \(\frac{H}{h}\):\[\frac{H}{h} = \frac{d(\tan(21^{\circ}) + \tan(52^{\circ}))}{d \tan(52^{\circ})} = \frac{\tan(21^{\circ}) + \tan(52^{\circ})}{\tan(52^{\circ})}.\]
07

Compute Values and Conclusion

Compute the values: \(\tan(21^{\circ}) \approx 0.3839\) and \(\tan(52^{\circ}) \approx 1.2799\). Substituting these values:\[\frac{\tan(21^{\circ}) + \tan(52^{\circ})}{\tan(52^{\circ})} = \frac{0.3839 + 1.2799}{1.2799} \approx \frac{1.6638}{1.2799} \approx 1.3009.\]Since \(1.3009\) is less than \(1.5\), your friend's claim is incorrect.

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Key Concepts

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

Angles of Elevation and Depression
Understanding angles of elevation and depression is like understanding how we visually perceive the height of an object from a specific vantage point.
If you look upwards to see the top of an object, you are using an **angle of elevation**. Conversely, when you look downwards, you are using an **angle of depression**.
For example, when you estimated your line of sight to be at a certain angle above the horizontal, that was the angle of elevation, while the angle downwards to see the base is viewed as the angle of depression.
  • The **angle of elevation** from horizontal to view an object upwards is useful when you want to measure height differences.
  • In the given exercise, the angle of elevation to the friend's building top was 21掳.
  • The **angle of depression** helps when you are positioned above the object of interest.
  • Here, the angle of depression to the base of the building was 52掳.
These angles help in establishing geometric relationships between the observer's position and the object being observed.
You can geometrically relate these angles with horizontal lines, which together create known angles that can be analyzed using trigonometry.
Tangent Function
The tangent function is a crucial trigonometric tool used in calculating heights and distances.
In right-angled triangles, the tangent of an angle is the ratio between the opposite side and the adjacent side.
Therefore, when you know the angle and one side, you can find the other. This makes the tangent function especially useful for problems involving angles of elevation and depression.
  • Mathematically, \( an(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}\).
  • In the exercise, the heights were calculated using this function.
  • For an angle \( \theta \ = 21^{掳} \), it relates the additional height (opposite) seen from the horizontal line of sight (adjacent) between buildings.
In simpler terms, the tangent function lets you unlock heights without having to measure every inch physically.
It helps you calculate the unknown parts of triangles when certain other measurements, like angles, are known.
Hence, you were able to apply it to find the height ratios between the buildings by using these angles.
Ratios in Geometry
Ratios are essential in understanding relationships between different lengths and sizes within geometry.
Specifically, ratios in trigonometry help compare one aspect of a geometric shape (like height) against another aspect.
  • In the exercise, finding the height ratio was crucial to proving or disproving the statement about the buildings' heights.
  • The previous steps allowed for forming expressions of heights using trigonometry for both buildings to finally set up a ratio.
  • The equation for the height ratio derived was \(\frac{H}{h} = \frac{\tan(21^{掳}) + \tan(52^{掳})}{\tan(52^{掳})}\).
This ratio compared how tall your friend's building is compared to yours, helping check if one is indeed 1.5 times taller.
By using standard trigonometric calculations, we found that it simply was not the case.
Ratios hence play a pivotal role, providing measurable comparisons to assist in drawing logical conclusions.

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Most popular questions from this chapter

A pilot flies her route in two straight-line segments. The displacement vector \(\overrightarrow{\mathbf{A}}\) for the first segment has a magnitude of \(244 \mathrm{km}\) and a direction \(30.0^{\circ}\) north of east. The displacement vector \(\overrightarrow{\mathbf{B}}\) for the second segment has a magnitude of \(175 \mathrm{km}\) and a direction due west. The resultant displacement vector is \(\overrightarrow{\mathbf{R}}=\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\) and makes an angle \(\theta\) with the direction due east. Using the component method, find the magnitude of \(\overrightarrow{\mathbf{R}}\) and the directional angle \(\theta\).

You and your team are exploring a river in South America when you come to the bottom of a tall waterfall. You estimate the cliff over which the water flows to be about 100 feet tall. You have to choose between climbing the cliff or backtracking and taking another route, but climbing the cliff would cut two hours off of your trip. There is only one experienced climber in the group: she would climb the cliff alone and drop a rope over the edge to lift supplies and allow the others to climb without packs. The climber estimates it will take her 45 minutes to get to the top. However, you are concerned that the rope might be too short to reach the bottom of the cliff (it is exactly \(30.0 \mathrm{m}\) long \() .\) If it is too short, she'll have to climb back down (another 45 minutes) and you will be too far behind schedule to get to your destination before dark. As you contemplate how to determine whether the rope is long enough, you notice that the late afternoon shadow of the cliff grows as the sun descends over its edge. You suddenly remember your trigonometry. You measure the length of the shadow from the base of the cliff to the shadow's edge (\(144 \mathrm{ft}\)), and the angle subtended between the base and top of the cliff measured from the shadow's edge. The angle is \(38.1^{\circ} .\) Do you send the climber, or start backtracking to take another route?

The components of vector \(\overrightarrow{\mathbf{A}}\) are \(A_{x}\) and \(A,\) (both positive), and the angle that it makes with respect to the positive \(x\) axis is \(\theta\). Find the angle \(\theta\) if the components of the displacement vector \(\overrightarrow{\mathbf{A}}\) are (a) \(A_{x}=12 \mathrm{m}\) and \(A_{y}=12 \mathrm{m},\) (b) \(A_{x}=17 \mathrm{m}\) and \(A_{y}=12 \mathrm{m},\) and (c) \(A_{x}=12 \mathrm{m}\) and \(A_{y}=17 \mathrm{m}\).

Vector \(\overrightarrow{\mathbf{A}}\) has a magnitude of 12.3 units and points due west. Vector \(\overrightarrow{\mathbf{B}}\) points due north. (a) What is the magnitude of \(\overrightarrow{\mathbf{B}}\) if \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\) has a magnitude of 15.0 units? (b) What is the direction of \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\) relative to due west? (c) What is the magnitude of \(\overrightarrow{\mathbf{B}}\) if \(\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}\) has a magnitude of 15.0 units? (d) What is the direction of \(\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}\) relative to due west?

An ocean liner leaves New York City and travels \(18.0^{\circ}\) north of east for \(155 \mathrm{km} .\) How far east and how far north has it gone? In other words, what are the magnitudes of the components of the ship's displacement vector in the directions (a) due east and (b) due north?
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