Math Problem Statement

From the top of a lighthouse, the angle of depression is 400 to a ship traveling away from the lighthouse. Twenty minutes later, the ship is spotted at an angle of depression of 100. How fast is the boat traveling per hour if the height of the lighthouse, from the water level, is 5km? (Show your work)

Solution

To solve this problem, we need to determine the speed of the ship by analyzing the change in distance over time. We'll use trigonometry, particularly the tangent function, which relates angles of depression to distances.

Given:

  • Height of the lighthouse, h=5kmh = 5 \, \text{km}
  • Initial angle of depression, θ1=40\theta_1 = 40^\circ
  • Angle of depression after 20 minutes, θ2=10\theta_2 = 10^\circ
  • Time interval, t=20minutest = 20 \, \text{minutes}

Step 1: Determine the initial and final distances of the ship from the lighthouse

Using the tangent function for the initial and final angles of depression: tan(θ1)=hd1\tan(\theta_1) = \frac{h}{d_1} tan(40)=5d1\tan(40^\circ) = \frac{5}{d_1} d1=5tan(40)d_1 = \frac{5}{\tan(40^\circ)}

For the second angle of depression: tan(θ2)=hd2\tan(\theta_2) = \frac{h}{d_2} tan(10)=5d2\tan(10^\circ) = \frac{5}{d_2} d2=5tan(10)d_2 = \frac{5}{\tan(10^\circ)}

Step 2: Calculate the distances

We can now calculate the distances using the tangent values.

d1=5tan(40)d_1 = \frac{5}{\tan(40^\circ)} d2=5tan(10)d_2 = \frac{5}{\tan(10^\circ)}

Let's calculate these values:

  • tan(40)0.8391\tan(40^\circ) \approx 0.8391
  • tan(10)0.1763\tan(10^\circ) \approx 0.1763

d1=50.83915.96kmd_1 = \frac{5}{0.8391} \approx 5.96 \, \text{km} d2=50.176328.37kmd_2 = \frac{5}{0.1763} \approx 28.37 \, \text{km}

Step 3: Determine the distance traveled by the ship

Distance traveled=d2d1=28.37km5.96km22.41km\text{Distance traveled} = d_2 - d_1 = 28.37 \, \text{km} - 5.96 \, \text{km} \approx 22.41 \, \text{km}

Step 4: Convert time from minutes to hours

t=20minutes=2060hours=13hourst = 20 \, \text{minutes} = \frac{20}{60} \, \text{hours} = \frac{1}{3} \, \text{hours}

Step 5: Calculate the speed of the ship

Speed=Distance traveledTime=22.41km13hours=22.41×3km/h67.23km/h\text{Speed} = \frac{\text{Distance traveled}}{\text{Time}} = \frac{22.41 \, \text{km}}{\frac{1}{3} \, \text{hours}} = 22.41 \times 3 \, \text{km/h} \approx 67.23 \, \text{km/h}

Final Answer:

The ship is traveling at approximately 67.23 km/h.

Would you like more details or have any questions?


Here are five related questions to explore further:

  1. How would the problem change if the lighthouse were only 2 km high?
  2. What would be the ship's speed if the angles of depression were 30° and 5° instead?
  3. How do changes in the angle of depression affect the distance calculation?
  4. Could this problem be solved using a different trigonometric function?
  5. What assumptions are made in this problem regarding the ship's movement?

Tip: Understanding how to use trigonometric functions like tangent to relate angles and distances can be very useful in solving real-world problems involving height and distance.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Geometry

Formulas

Tangent function: \( \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} \)

Theorems

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Suitable Grade Level

High School