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)

To solve this problem, we need to use trigonometry to relate the angle of depression and the height of the lighthouse to the distance of the ship from the lighthouse. Let's break it down step by step.

### Given:
- Height of the lighthouse, \( h = 5 \) km
- Initial angle of depression, \( \theta_1 = 40^\circ \)
- Final angle of depression, \( \theta_2 = 10^\circ \)
- Time difference between the two sightings, \( t = 20 \) minutes

### Step 1: Find the initial distance of the ship from the lighthouse

Let \( d_1 \) be the initial distance of the ship from the base of the lighthouse when the angle of depression is \( \theta_1 = 40^\circ \).

Using the tangent function:

\[
\tan(40^\circ) = \frac{h}{d_1}
\]

\[
d_1 = \frac{h}{\tan(40^\circ)}
\]

### Step 2: Find the final distance of the ship from the lighthouse

Let \( d_2 \) be the distance of the ship from the base of the lighthouse when the angle of depression is \( \theta_2 = 10^\circ \).

Using the tangent function:

\[
\tan(10^\circ) = \frac{h}{d_2}
\]

\[
d_2 = \frac{h}{\tan(10^\circ)}
\]

### Step 3: Calculate the distances

Now, let's calculate \( d_1 \) and \( d_2 \):

\[
d_1 = \frac{5 \text{ km}}{\tan(40^\circ)}
\]
\[
d_2 = \frac{5 \text{ km}}{\tan(10^\circ)}
\]

We can plug in the values for \( \tan(40^\circ) \) and \( \tan(10^\circ) \):

\[
d_1 \approx \frac{5 \text{ km}}{0.8391} \approx 5.96 \text{ km}
\]
\[
d_2 \approx \frac{5 \text{ km}}{0.1763} \approx 28.37 \text{ km}
\]

### Step 4: Find the distance traveled by the ship

The distance traveled by the ship in 20 minutes is:

\[
\text{Distance traveled} = d_2 - d_1 \approx 28.37 \text{ km} - 5.96 \text{ km} \approx 22.41 \text{ km}
\]

### Step 5: Find the speed of the ship

The time taken is 20 minutes, which is \( \frac{20}{60} \) hours, or \( \frac{1}{3} \) hours.

The speed of the ship \( v \) is given by:

\[
v = \frac{\text{Distance traveled}}{\text{Time taken}} = \frac{22.41 \text{ km}}{\frac{1}{3} \text{ hours}} \approx 67.23 \text{ km/h}
\]

### Final Answer:

The ship is traveling at approximately **67.23 km/h**.

Would you like to see more details, or do you have any questions?

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Here are five related questions you might find interesting:

1. How would the result change if the lighthouse were taller or shorter?
2. What if the ship's speed was not constant over the 20 minutes?
3. How would you calculate the speed if the angles were given in radians instead of degrees?
4. What trigonometric identities are essential for solving problems like this?
5. How can you use calculus to model the ship's distance over time?

**Tip:** In problems involving angles of depression or elevation, always draw a diagram to help visualize the relationships between angles, heights, and distances.

Learn how to calculate the speed of a boat using trigonometry with an angle of depression problem. This problem involves finding distances using the tangent function and calculating speed based on time and distance. Suitable for Grades 10-12 students studying trigonometry and applied mathematics.

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