This article provides practice problems involving angles of elevation and depression, focusing on examples that would typically be found in an 8th-5th grade level math curriculum. The problems increase in difficulty, allowing for a gradual understanding of the concepts.
Understanding Angles of Elevation and Depression
Before we start, let's refresh our understanding:
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Angle of Elevation: This is the angle formed between the horizontal line of sight and the line of sight up to an object. Imagine you are looking up at a bird; the angle between your horizontal gaze and your gaze at the bird is the angle of elevation.
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Angle of Depression: This is the angle formed between the horizontal line of sight and the line of sight down to an object. Imagine you are standing on a cliff looking down at a boat; the angle between your horizontal gaze and your gaze at the boat is the angle of depression.
Important Note: The angle of elevation from point A to point B is equal to the angle of depression from point B to point A.
Practice Problems
Here are some practice problems involving angles of elevation and depression:
Problem 1: The Flagpole
A flagpole casts a shadow 12 meters long. The angle of elevation from the tip of the shadow to the top of the flagpole is 30 degrees. How tall is the flagpole? (Assume the flagpole is perpendicular to the ground).
Solution:
This problem uses basic trigonometry (SOH CAH TOA). We can use the tangent function:
tan(30°) = opposite / adjacent = height of flagpole / length of shadow
tan(30°) = height / 12 meters
Solving for the height:
height = 12 meters * tan(30°) ≈ 6.93 meters
Problem 2: The Airplane
An airplane is flying at an altitude of 5000 meters. The angle of depression from the airplane to a point on the ground is 15 degrees. How far is the point on the ground from a point directly below the airplane?
Solution:
Again, we use trigonometry. The angle of depression from the airplane to the ground is equal to the angle of elevation from the ground to the airplane. We use the tangent function:
tan(15°) = opposite / adjacent = altitude / distance
tan(15°) = 5000 meters / distance
Solving for the distance:
distance = 5000 meters / tan(15°) ≈ 18,660 meters
Problem 3: The Mountain Climber
A mountain climber is standing at the base of a mountain. They measure the angle of elevation to the top of the mountain to be 40 degrees. They walk 100 meters closer to the mountain and measure the angle of elevation again; this time it's 50 degrees. How tall is the mountain?
Solution:
This problem requires setting up two trigonometric equations using tangent and solving a system of equations. This is a more challenging problem that might be appropriate for the upper end of the target grade range. Let 'h' represent the height of the mountain and 'x' represent the initial distance to the mountain. We get two equations:
tan(40°) = h / x tan(50°) = h / (x - 100)
Solving this system of equations (typically through substitution) will give the value of 'h', the height of the mountain.
Further Practice
These problems offer a foundation for understanding angles of elevation and depression. To further improve your skills, try to find additional problems involving similar triangles and word problems applying these concepts in different real-world scenarios. Remember to always draw a diagram to visualize the problem.
