Problem 17
Question
Weather radar is capable of measuring both the angle of elevation to the top of a thunderstorm and its range (the horizontal distance to the storm). If the range of a storm is \(90 \mathrm{~km}\) and the angle of elevation is \(4^{\circ},\) can a passenger plane that is able to climb to \(10 \mathrm{~km}\) fly over the storm?
Step-by-Step Solution
Verified Answer
Yes, the plane can fly over the storm, as its maximum altitude is higher than the storm height.
1Step 1: Identify Right Triangle Components
In this problem, we are dealing with a right triangle where the range to the storm forms the base. The top of the storm corresponds to the vertical side, and the angle of elevation is given as \(4^{\circ}\). We need to find the height (opposite side) of the right triangle using these components.
2Step 2: Use Trigonometric Function to Find Height
We can use the tangent function, which relates the angle of elevation to the opposite side (height of the storm, \(h\)) and the adjacent side (range of the storm): \[ \tan(\theta) = \frac{h}{\text{range}} \] Substituting the known values: \[ \tan(4^{\circ}) = \frac{h}{90} \]
3Step 3: Solve for Height
Solve the equation for \(h\) (height of the storm): \[ h = 90 \times \tan(4^{\circ}) \] Calculate \(h\) using a calculator to find \( \tan(4^{\circ}) \): \( h \approx 90 \times 0.0699 \approx 6.291 \text{ km} \).
4Step 4: Compare Height to Maximum Altitude of Plane
The height of the storm top is approximately \(6.291 \text{ km}\). A passenger plane can climb to \(10 \text{ km}\). Since \(10 \text{ km} > 6.291 \text{ km}\), the plane can fly over the storm.
Key Concepts
Right TriangleAngle of ElevationTangent Function
Right Triangle
A right triangle is a fundamental concept in trigonometry. It consists of three sides and includes a right angle, which is exactly 90 degrees. The side opposite the right angle is the longest side and is called the hypotenuse. The other two sides are known as the legs of the triangle. These legs include:
- Adjacent Side: It is the side that touches the angle in question, aside from the hypotenuse.
- Opposite Side: It lies across from the angle in question and does not touch it.
Angle of Elevation
The angle of elevation is a key element in the study of trigonometry and specifically in problems involving height and distance. It is defined as the angle between the horizontal line and the line of sight looking up towards an object. This angle indicates how high one would need to tilt their view from the horizontal to see the top of an object.
Understanding the angle of elevation can help in various applications such as navigation, aviation, and construction. For example, when determining the height of tall structures or natural formations, knowing the angle of elevation can be vital.
In our thunderstorm problem, the angle of elevation was 4 degrees. This is the angle formed between the level ground (horizontal line) and the observer's line of sight to the storm's top. This knowledge allows us to deploy trigonometric functions to find unknown heights, which is crucial for determining whether an airplane can safely fly over obstacles.
Tangent Function
The tangent function is a fundamental trigonometric function that relates the angles and sides of a right triangle to one another. Specifically, it relates the ratio of the opposite side to the adjacent side. The mathematical formula for the tangent of an angle \( \theta \) is:\[ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} \]In practical contexts, such as estimating the height of a thunderstorm, the tangent function stands out because it directly connects the height we want to find with known measures like the range and angle of elevation. By rearranging the tangent formula, we can solve for the unknown height:\[ \text{Opposite} = \text{Adjacent} \times \tan(\theta) \]For our specific problem, substituting the given values allowed us to compute the storm's height by using the formula \( 90 \times \tan(4^{\circ}) \), resulting in approximately 6.291 km. This demonstrates how the tangent function is applied in real-world scenarios to solve height measurement problems and ensure safety, such as confirming that an aircraft can surpass the altitude of a storm.
Other exercises in this chapter
Problem 17
Find \(\mathbf{u}-4 \mathbf{v}\) and \(2 \mathbf{u}+5 \mathbf{v}\). $$ \mathbf{u}=\frac{1}{2} \mathbf{i}-\frac{3}{2} \mathbf{j}, \mathbf{v}=2 \mathbf{i} $$
View solution Problem 17
A ship sails due west from a harbor for 22 nautical miles. It then sails \(\mathrm{S} 62^{\circ} \mathrm{W}\) for another 15 nautical miles. How far is the ship
View solution Problem 17
Use the Law of Sines to solve the triangle. $$ \beta=30^{\circ}, a=10, b=7 $$
View solution Problem 18
Find \(\mathbf{u}-4 \mathbf{v}\) and \(2 \mathbf{u}+5 \mathbf{v}\). $$ \mathbf{u}=2 \mathbf{i}-3 \mathbf{j}, \mathbf{v}=3 \mathbf{i}-2 \mathbf{j} $$
View solution