Understanding the Problem: Calculating Boat Speed using Trigonometry

This problem involves using trigonometry, specifically the concept of angles of depression, to find the speed of a moving boat. We are given the height of a cliff from which the observation is made, and the change in the angle of depression of a boat over a specific time interval. We need to determine the speed of the boat in metres per hour.

The angle of depression is the angle between the horizontal line of sight and the line of sight downwards to an object. The angle of depression from the top of the cliff to the boat is equal to the angle of elevation from the boat to the top of the cliff, due to alternate interior angles being equal.

Setting up the Geometry for Boat Observation

Let's visualize the scenario. We have a right-angled triangle formed by the cliff (vertical side), the horizontal distance from the base of the cliff to the boat (horizontal side), and the line of sight from the top of the cliff to the boat (hypotenuse). The height of the cliff is given as 150 m.

Let:

• \(H\) be the height of the cliff = 150 m.
• \(\theta_1\) be the initial angle of depression = 60°.
• \(\theta_2\) be the final angle of depression = 45°.
• \(d_1\) 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.

Since the boat is observed to change its angle of depression from 60° to 45°, it means the boat is moving away from the base of the cliff. The angle of elevation from the boat to the top of the cliff will change from 60° to 45°.

Calculating Distances using Trigonometry (Tangent Function)

We can use the tangent function, which relates the opposite side (cliff height) to the adjacent side (distance from the base) in a right-angled triangle:

\(\tan(\text{angle of elevation}) = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{\text{Height of cliff}}{\text{Distance from base}}\)

Initial Observation (Angle 60°)

When the angle of depression is 60°, the angle of elevation from the boat is also 60°.

\(\tan 60^\circ = \frac{H}{d_1}\)

We know \(H = 150\) m and \(\tan 60^\circ = \sqrt{3}\). So,

\(\sqrt{3} = \frac{150}{d_1}\)

\(d_1 = \frac{150}{\sqrt{3}}\) metres.

Final Observation (Angle 45°)

When the angle of depression is 45°, the angle of elevation from the boat is also 45°.

\(\tan 45^\circ = \frac{H}{d_2}\)

We know \(H = 150\) m and \(\tan 45^\circ = 1\). So,

\(1 = \frac{150}{d_2}\)

\(d_2 = 150\) metres.

Calculating the Distance Moved by the Boat

The boat moved from an initial distance \(d_1\) to a final distance \(d_2\) away from the base of the cliff. The distance moved by the boat is the difference between these two distances:

Distance moved, \(D = d_2 - d_1\)

\(D = 150 - \frac{150}{\sqrt{3}}\) metres.

Calculating the Speed of the Boat

The boat covered the distance \(D\) in a time interval of 2 minutes. We need to calculate the speed in metres per hour. First, convert the time into hours:

Time = 2 minutes = \(\frac{2}{60}\) hours = \(\frac{1}{30}\) hours.

Speed is defined as the distance covered per unit of time:

Speed = \(\frac{\text{Distance moved}}{\text{Time taken}}\)

Speed = \(\frac{150 - \frac{150}{\sqrt{3}}}{\frac{1}{30}}\) metres per hour.

Speed = \(30 \times \left(150 - \frac{150}{\sqrt{3}}\right)\) metres per hour.

Factor out 150 from the expression in the parenthesis:

Speed = \(30 \times 150 \left(1 - \frac{1}{\sqrt{3}}\right)\) metres per hour.

Speed = \(4500 \left(1 - \frac{1}{\sqrt{3}}\right)\) metres per hour.

To match the format of the options, combine the terms inside the parenthesis by finding a common denominator:

Speed = \(4500 \left(\frac{\sqrt{3}}{{\sqrt{3}}} - \frac{1}{\sqrt{3}}\right)\) metres per hour.

Speed = \(4500 \left(\frac{{\sqrt{3}} - 1}{\sqrt{3}}\right)\) metres per hour.

Conclusion

The speed of the boat is \(\frac{{4500\left( {\sqrt 3 - 1} \right)}}{{\sqrt 3 }}\) metres per hour.

Revision Table: Key Concepts for Heights and Distances

Additional Information: Applied Trigonometry

Problems involving angles of elevation and depression are common applications of trigonometry. They help us find heights and distances of objects that are difficult to measure directly. These concepts are used in surveying, navigation, engineering, and many other fields.

Key steps in solving such problems often include:

• Drawing a clear diagram representing the situation.
• Identifying the right-angled triangles formed.
• Using the given angles (elevation or depression, remembering the relationship between them) and known side lengths.
• Applying appropriate trigonometric ratios (\(\sin\), \(\cos\), or \(\tan\)) to find unknown lengths.
• Solving the equations derived from the trigonometric ratios.
• If the problem involves motion (like the moving boat), calculate the distance covered and use the speed = distance/time formula, ensuring consistent units.

Remembering the standard values for trigonometric ratios of common angles like 0°, 30°, 45°, 60°, and 90° is often helpful in solving these problems quickly.