Question

Two small boats are sailing in the sea on the two sides of a big ship. The angle of elevation of the top of the big ship is observed from the boat are 30 degree and 45 degree. If the height of the big ship is 200 meters as measured from the surface of the sea, calculate the distance between two small boats?

• A. 546.41 m
• B. 545.41 m
• C. 545.72 m
• D. 545.67 m

Hint:

In trigonometry problems with standard angles, remember these quick relations for a right triangle with height H:

If angle = 45°, Base = Height.
If angle = 30°, Base = Height \(\times \sqrt{3}\).
If angle = 60°, Base = Height / \(\sqrt{3}\).

Answer 1

The Correct Option is A

Step 1: Understanding the Problem
This is a heights and distances problem in trigonometry. We have a ship of a certain height, and two boats on opposite sides. This forms two right-angled triangles with a common height. We need to find the total distance between the boats.
Let the height of the ship (AB) be \(h = 200\) m.
Let the two boats be at points C and D.
Let the distance of the first boat from the base of the ship be \(x = BC\). Angle of elevation = 45°.
Let the distance of the second boat from the base of the ship be \(y = BD\). Angle of elevation = 30°.
We need to find the total distance \(CD = x + y\).

Step 2: Key Formula or Approach
We will use the tangent ratio: \( \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} \).

Step 3: Detailed Explanation
1. Calculate distance x (for the 45° angle):
In the right-angled triangle ABC: \[ \tan(45^\circ) = \frac{AB}{BC} = \frac{200}{x} \] Since \( \tan(45^\circ) = 1 \): \[ 1 = \frac{200}{x} \] \[ x = 200 \text{ m} \] 2. Calculate distance y (for the 30° angle):
In the right-angled triangle ABD: \[ \tan(30^\circ) = \frac{AB}{BD} = \frac{200}{y} \] Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \): \[ \frac{1}{\sqrt{3}} = \frac{200}{y} \] \[ y = 200\sqrt{3} \text{ m} \] 3. Calculate the total distance:
The total distance between the boats is \( x + y \). \[ \text{Total Distance} = 200 + 200\sqrt{3} = 200(1 + \sqrt{3}) \] Using the value \( \sqrt{3} \approx 1.732 \): \[ \text{Total Distance} = 200(1 + 1.732) = 200(2.732) = 546.4 \text{ m} \] Looking at the options, 546.41 m is the closest.

Step 4: Final Answer
The distance between the two small boats is approximately 546.41 m. Therefore, option (A) is the correct answer.