Problem 26
Question
From a point 15 meters above level ground, a surveyor measures the angle of depression of an object on the ground at \(68^{\circ}\). Approximate the distance from the object to the point on the ground directly beneath the surveyor.
Step-by-Step Solution
Verified Answer
The distance is approximately 6.06 meters.
1Step 1: Understanding the Problem
We need to find the horizontal distance from the object to the base of a point where the surveyor is situated, which is 15 meters above the ground. The angle of depression from the surveyor to the object on the ground is given as \(68^{\circ}\).
2Step 2: Setting Up the Triangle
Imagine a right triangle formed by the horizontal line from the object to the point directly beneath the surveyor (ground level), the vertical line from the surveyor to the ground (15 meters), and the line of sight from the surveyor to the object. The angle of depression \(68^{\circ}\) is equal to the angle of elevation from the object to the surveyor's line of sight.
3Step 3: Using Trigonometry
Use the tangent function, which relates the opposite side (15 meters) to the adjacent side (the distance we need to find), using the angle of elevation or depression. The tangent function is given by \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \).
4Step 4: Applying the Tangent Function
Since \( \theta = 68^{\circ} \), apply the formula: \[ \tan(68^{\circ}) = \frac{15}{d} \] where \(d\) is the distance we want to find.
5Step 5: Solving for Distance
Rearrange the equation to solve for \(d\): \[ d = \frac{15}{\tan(68^{\circ})} \].
6Step 6: Calculating the Result
Calculate \(d\) using a calculator: \[ \tan(68^{\circ}) \approx 2.475 \]. Thus, \[ d = \frac{15}{2.475} \approx 6.06 \]. So, the distance is approximately 6.06 meters.
Key Concepts
Angle of DepressionTangent FunctionTrigonometric Ratios
Angle of Depression
In trigonometry, the angle of depression is defined as the angle formed by the horizontal line and the line of sight from an observer to an object below them. This angle is crucial in solving problems involving heights and distances, especially when determining unseen dimensions.
Let's break it down: Imagine you are standing at a high point, looking down at an object on the ground. The line of sight from your eyes to the object forms the hypotenuse of a right triangle, while the horizontal line is the base of this triangle. The angle of depression is the angle formed between this horizontal line and your line of sight.
Why is it termed as 'depression'? Because it indicates a downward direction from a higher point to a lower point, just like feeling a downward mood can be depressing. Keep this imagery in mind to easily recall this concept!
In practical problems, the angle of depression is equal to the angle of elevation from the object on the ground to the observer at a height. This is because horizontal lines are parallel to each other, reflecting principles from geometry.
Let's break it down: Imagine you are standing at a high point, looking down at an object on the ground. The line of sight from your eyes to the object forms the hypotenuse of a right triangle, while the horizontal line is the base of this triangle. The angle of depression is the angle formed between this horizontal line and your line of sight.
Why is it termed as 'depression'? Because it indicates a downward direction from a higher point to a lower point, just like feeling a downward mood can be depressing. Keep this imagery in mind to easily recall this concept!
In practical problems, the angle of depression is equal to the angle of elevation from the object on the ground to the observer at a height. This is because horizontal lines are parallel to each other, reflecting principles from geometry.
Tangent Function
The tangent function is an essential part of trigonometry, relating the angles and sides of a right triangle. Specifically, the tangent of an angle in a right triangle is the ratio of the length of the side opposite to the angle to the length of the side adjacent to the angle.
This can be understood through the mathematical expression:
The tangent function is widely used because, with the knowledge of an angle and one side, you can find the other side. In this exercise, the surveyor used the known height (15 meters) as the opposite side and the angle of depression (68 degrees) to find the unknown horizontal distance.
This relationship is why trigonometry is a powerful tool in solving real-world problems.
This can be understood through the mathematical expression:
- \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \)
The tangent function is widely used because, with the knowledge of an angle and one side, you can find the other side. In this exercise, the surveyor used the known height (15 meters) as the opposite side and the angle of depression (68 degrees) to find the unknown horizontal distance.
This relationship is why trigonometry is a powerful tool in solving real-world problems.
Trigonometric Ratios
Trigonometric ratios refer to the relations of the sides of right triangles concerning their angles. The primary trigonometric ratios are sine, cosine, and tangent, which each use different parts of the triangle.
To understand these ratios:
Understanding and memorizing these relationships through practice will help greatly when you're required to solve problems involving distances and angles, just like the exercise above.
To understand these ratios:
- Sine ( \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \) ) is the ratio of the length of the opposite side to the length of the hypotenuse.
- Cosine ( \( \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \) ) is the ratio of the length of the adjacent side to the hypotenuse.
- Tangent ( \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \) ) is the ratio of the opposite side to the adjacent side.
Understanding and memorizing these relationships through practice will help greatly when you're required to solve problems involving distances and angles, just like the exercise above.
Other exercises in this chapter
Problem 25
Exer. \(25-28:\) Express the angle in terms of degrees, minutes, and seconds, to the nearest second. $$63.169^{\circ}$$
View solution Problem 26
Verify the identity by transforming the left hand side into the right-hand side. $$\cot (-x) \cos (-x)+\sin (-x)=-\csc x$$
View solution Problem 26
Find the amplitude, the period, and the phase shift and sketch the graph of the equation. \(y=\frac{1}{2} \cos \frac{\pi}{2} x\)
View solution Problem 26
Find the period and sketch the graph of the equation. Show the asymptotes. $$y=-\frac{1}{3} \cot (3 x-\pi)$$
View solution