Question

The angle of depression of the two ships from the top of a lighthouse are \( 30^\circ \) and \( 45^\circ \). If one ship is exactly behind the other on the same side of the lighthouse and are 50 metres apart from each other, find the height of the lighthouse from the sea level.

Hint:

When solving problems involving angles of depression, use the tangent function to relate the height of the object and the distance to the observer.

Answer 1

Let the height of the lighthouse be \( h \) metres. Let the distance between the lighthouse and the first ship be \( x_1 \), and the distance between the lighthouse and the second ship be \( x_2 \). The distance between the two ships is given as 50 metres, so: \[ x_2 - x_1 = 50. \] Triangle 1: For the first ship From the angle of depression \( 30^\circ \) and using the tangent function, we get: \[ \tan(30^\circ) = \frac{h}{x_1}. \] Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we have: \[ \frac{1}{\sqrt{3}} = \frac{h}{x_1} \quad \Rightarrow \quad x_1 = \sqrt{3}h. \] Triangle 2: For the second ship From the angle of depression \( 45^\circ \) and using the tangent function again, we get: \[ \tan(45^\circ) = \frac{h}{x_2}. \] Since \( \tan(45^\circ) = 1 \), we have: \[ 1 = \frac{h}{x_2} \quad \Rightarrow \quad x_2 = h. \] Set up the equation for the distance between the ships Now, substitute \( x_1 = \sqrt{3}h \) and \( x_2 = h \) into the equation \( x_2 - x_1 = 50 \): \[ h - \sqrt{3}h = 50. \] Factor out \( h \): \[ h(1 - \sqrt{3}) = 50. \] Solve for \( h \): \[ h = \frac{50}{1 - \sqrt{3}}. \] Multiply both the numerator and denominator by \( 1 + \sqrt{3} \) to rationalize the denominator: \[ h = \frac{50(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} = \frac{50(1 + \sqrt{3})}{1 - 3} = \frac{50(1 + \sqrt{3})}{-2}. \] Thus: \[ h = -25(1 + \sqrt{3}) \, \text{metres}. \] Since the height of the lighthouse cannot be negative, we take the positive value, and the height of the lighthouse is: \[ h = 25(1 + \sqrt{3}) \, \text{metres}. \]
Conclusion: The height of the lighthouse is \( 25(1 + \sqrt{3}) \, \text{metres}. \)