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An observer, whose eyes are \(1.83 \mathrm{~m}\) above the ground, is standing \(32.0 \mathrm{~m}\) away from a tree. The ground is level, and the tree is growing perpendicular to it. The observer's line of sight with the treetop makes an angle of \(20.0^{\circ}\) above the horizontal. How tall is the tree?

Short Answer

Expert verified
The tree is approximately 13.48 meters tall.

Step by step solution

01

Understand the Right Triangle

Visualize the problem as a right triangle where the observer's position, the top of the tree, and the base of the tree form the vertices. The observer's line of sight to the treetop represents the hypotenuse, the distance from the observer to the base of the tree is one leg, and the vertical difference in height (the unknown) is the other leg.
02

Identify Known Values

The distance from the observer to the tree is 32.0 m, the angle of elevation is 20.0 degrees, and the height of the observer's eyes is 1.83 m. These will be used to find the tree's height.
03

Use the Tangent Function

Apply the tangent trigonometric function which relates the angle of elevation to the opposite side (extra height of tree above observer's eye level) and the adjacent side (distance from the observer to the tree): \[\tan(20.0^{\circ}) = \frac{\text{Height above eye level}}{32.0}\]
04

Solve for Height Above Eye Level

Rearrange the equation to solve for the height of the tree above the observer's eye level:\[\text{Height above eye level} = 32.0 \times \tan(20.0^{\circ})\] Calculate this value using a calculator.
05

Compute the Total Height of the Tree

Add the observer's eye level height of 1.83 m to the height above eye level to find the total height of the tree:\[\text{Total tree height} = 1.83 + 32.0 \times \tan(20.0^{\circ})\]
06

Calculate and Verify

Use the value calculated in Step 4, add the observer's eye level height to it, and verify: \[\text{Total height of the tree} = 1.83 + (32.0 \times 0.3640) \approx 1.83 + 11.65 = 13.48 \, \text{m}\]

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Key Concepts

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

Right Triangle
In the world of geometry, a right triangle is incredibly useful. It's defined by one angle being exactly 90 degrees. This is the type of triangle everyone learns about in basic geometry classes. Picture the scenario in the original exercise: the observer stands on one vertex of the triangle, the top of the tree makes another vertex, and the base of the tree forms the third.
A right triangle comprises three sides:
  • The hypotenuse is the longest side and in our problem, it is the observer's line of sight to the treetop.
  • The two legs are the other sides. One leg is the horizontal distance from the observer to the tree, which is given as 32 meters.
  • The other leg represents the vertical difference in height—the extra height of the tree above the observer’s eye level.
Understanding right triangles allows us to find unknown lengths when we have an angle and a side, thanks to trigonometry.
Angle of Elevation
The angle of elevation is a specific type of angle measurement used in geometry. It is the angle formed from the horizontal up to a line of sight. Imagine you're standing on level ground and looking up at something—like the top of a tree in this scenario. The angle your line of sight makes with the horizontal is the angle of elevation.
In this exercise, the observer has an angle of elevation of 20 degrees when gazing at the top of the tree.
Knowing this angle is crucial because angles allow us to use trigonometric functions to solve for unknown distances or heights. Angles of elevation are encountered often in real-world situations, such as determining the height of buildings, towers, or even mountains without having to climb them!
Tangent Function
The tangent function is a fundamental element of trigonometry, especially concerning right triangles. In the context of this exercise, the tangent of an angle in a right triangle links the ratio of the Opposite side to the Adjacent side. For this exercise, the opposite side is the height difference above the observer's eye level, and the adjacent side is the horizontal distance from the observer to the tree.
Mathematically, the tangent function is given by: \[\tan(\Theta) = \frac{\text{Opposite}}{\text{Adjacent}}\]Where \(\Theta\) is the angle of elevation. Here it is \(20^{\circ}\). This gives us:\[\tan(20.0^{\circ}) = \frac{\text{Height above eye level}}{32.0}\]Using a calculator, you find the tangent of 20 degrees (approximately 0.364), which helps discover the height above eye level when multiplied by the distance from the observer to the tree. The tangent function simplification provides a swift path to determining unknown dimensions in geometric configurations.

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Most popular questions from this chapter

The \(x\) vector component of a displacement vector \(\overrightarrow{\mathbf{r}}\) has a magnitude of \(125 \mathrm{~m}\) and points along the negative \(x\) axis. The \(y\) vector component has a magnitude of \(184 \mathrm{~m}\) and points along the negative \(y\) axis. Find the magnitude and direction of \(\overrightarrow{\mathbf{r}}\). Specify the direction with respect to the negative \(x\) axis.

(a) Considering the fact that \(3.28 \mathrm{ft}=1 \mathrm{~m}\), which is the larger unit for measuring area, \(1 \mathrm{ft}^{2}\) or \(1 \mathrm{~m}^{2} ?\) (b) Consider a \(1330-\mathrm{ft}^{2}\) apartment. With your answer to part (a) in mind and without doing any calculations, decide whether this apartment has an area that is greater than or less than \(1330 \mathrm{~m}^{2}\). In a \(1330-\mathrm{ft}^{2}\) apartment, how many square meters of area are there? Be sure that your answer is consistent with your answers to the Concept Questions.

The displacement vector \(\overrightarrow{\mathrm{A}}\) has scalar components of \(A_{x}=80.0 \mathrm{~m}\) and \(A_{y}=60.0 \mathrm{~m}\) The displacement vector \(\overrightarrow{\mathbf{B}}\) has a scalar component of \(B_{x}=60.0 \mathrm{~m}\) and a magnitude of \(B=75.0 \mathrm{~m} .\) The displacement vector \(\overrightarrow{\mathrm{C}}\) has a magnitude of \(C=100.0 \mathrm{~m}\) and is directed at an angle of \(36.9^{\circ}\) above the \(+x\) axis. Two of these vectors are equal. Determine which two, and support your choice with a calculation.

Vector \(\vec{A}\) has a magnitude of \(6.00\) units and points due east. Vector \(\overrightarrow{\mathbf{B}}\) points due north. (a) What is the magnitude of \(\overrightarrow{\mathrm{B}}\), if the vector \(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}\) points \(60.0^{\circ}\) north of east? (b) Find the magnitude of \(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}\).

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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