a moving boat is observed from the top of a 150m high cliff moving away from the cliff. the angle of depression of the boat changes from 60degree to 45 degree in 2 minutes. find the speed of the boat in m\h

To solve this problem, we will use trigonometry to relate the angles of depression to the horizontal distances of the boat from the base of the cliff. Then we'll find the speed of the boat.

### Step 1: Define the Problem
- Height of the cliff, \( h = 150 \) meters.
- Initial angle of depression, \( \theta_1 = 60^\circ \).
- Final angle of depression, \( \theta_2 = 45^\circ \).
- Time taken for the change in angle, \( t = 2 \) minutes.

Let:
- \( d_1 \) be the initial distance of the boat from the base of the cliff.
- \( d_2 \) be the final distance of the boat from the base of the cliff.

### Step 2: Use Trigonometry to Find Distances
We know that the angle of depression from the top of the cliff to the boat forms a right triangle, with the height of the cliff as the opposite side and the distance from the base of the cliff to the boat as the adjacent side.

For the initial position:
\[
\tan(\theta_1) = \frac{h}{d_1}
\]
\[
\tan(60^\circ) = \frac{150}{d_1}
\]
\[
\sqrt{3} = \frac{150}{d_1}
\]
\[
d_1 = \frac{150}{\sqrt{3}} = 50\sqrt{3} \, \text{meters}
\]

For the final position:
\[
\tan(\theta_2) = \frac{h}{d_2}
\]
\[
\tan(45^\circ) = \frac{150}{d_2}
\]
\[
1 = \frac{150}{d_2}
\]
\[
d_2 = 150 \, \text{meters}
\]

### Step 3: Calculate the Distance Traveled by the Boat
The distance traveled by the boat is:
\[
\Delta d = d_2 - d_1 = 150 - 50\sqrt{3} \, \text{meters}
\]

### Step 4: Calculate the Speed of the Boat
The time taken is 2 minutes, which is \( \frac{2}{60} \) hours.

The speed \( v \) of the boat is given by:
\[
v = \frac{\Delta d}{\text{time}} = \frac{150 - 50\sqrt{3}}{\frac{2}{60}} \, \text{m/h}
\]
\[
v = (150 - 50\sqrt{3}) \times \frac{60}{2} \, \text{m/h}
\]
\[
v = (150 - 50\sqrt{3}) \times 30 \, \text{m/h}
\]

### Step 5: Final Calculation
Let's calculate the exact speed of the boat.

\[
v = 30 \times (150 - 50\sqrt{3}) \, \text{m/h}
\]
\[
v = 4500 - 1500\sqrt{3} \, \text{m/h}
\]

Given \( \sqrt{3} \approx 1.732 \), the final speed is approximately:
\[
v \approx 4500 - 1500 \times 1.732 \, \text{m/h}
\]
\[
v \approx 4500 - 2598 \, \text{m/h}
\]
\[
v \approx 1902 \, \text{m/h}
\]

So, the speed of the boat is approximately **1902 meters per hour**.

Would you like further details or have any questions? Here are some related questions you might find interesting:

1. How would the result change if the time taken was 3 minutes instead of 2 minutes?
2. What would the speed be if the cliff height was 200 meters?
3. How does the angle of depression affect the distance calculation?
4. Can you derive the general formula for the speed of the boat using the given variables?
5. How would the solution change if the boat were approaching the cliff instead of moving away?

**Tip:** When solving problems involving angles of depression or elevation, always visualize the right triangle formed to relate the angles and distances using trigonometric functions.

Learn how to calculate the speed of a boat moving away from a 150-meter cliff using trigonometry. Solve for the boat's speed given changes in angle of depression over a 2-minute interval. Detailed step-by-step solution provided.

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