Finding Light House Height Using Angles of Depression

This problem involves using trigonometry to find the height of a light house. We are given the angles of depression from the top of the light house to two ships on opposite sides and collinear with the light house, as well as the distance between the ships.

Let's define the terms:

• Light House: A tall structure, represented as a vertical line.
• Ships: Located on the sea level, represented as points on a horizontal line.
• Angles of Depression: The angle between the horizontal line from the observer's eye (top of the light house) and the line of sight to an object below the horizontal line (the ships).

When considering angles of depression from the top of the light house to the ships, the angle of elevation from the ships to the top of the light house is equal to the angle of depression (alternate interior angles are equal when a transversal line intersects parallel lines - the horizontal line from the top of the light house is parallel to the sea level).

Let:

• \(H\) be the height of the light house (what we need to find).
• The light house be represented by the vertical line segment AB, where A is the top and B is the base.
• The two ships be C and D, located on opposite sides of the base B, such that C, B, and D are collinear on the sea level.

The angles of depression are given as 30° and 60°. Therefore, the angles of elevation from the ships to the top of the light house are also 30° and 60°.

• Angle of elevation from ship C to A is \( \angle ACB = 30^\circ \) (or 60°).
• Angle of elevation from ship D to A is \( \angle ADB = 60^\circ \) (or 30°).

Since the angles are different, the ships are at different distances from the base. Let's assume the ship with the 60° angle of elevation is closer to the base.

• Let \( \angle ADB = 60^\circ \) and \( \angle ACB = 30^\circ \).
• Let the distance from ship D to the base B be \( BD = x_1 \).
• Let the distance from ship C to the base B be \( BC = x_2 \).

We have two right-angled triangles: \(\triangle ABD\) and \(\triangle ABC\).

In \(\triangle ABD\):

\( \tan(60^\circ) = \frac{AB}{BD} = \frac{H}{x_1} \)

We know \( \tan(60^\circ) = \sqrt{3} \). So, \( \sqrt{3} = \frac{H}{x_1} \). This gives \( x_1 = \frac{H}{\sqrt{3}} \).

In \(\triangle ABC\):

\( \tan(30^\circ) = \frac{AB}{BC} = \frac{H}{x_2} \)

We know \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \). So, \( \frac{1}{\sqrt{3}} = \frac{H}{x_2} \). This gives \( x_2 = H\sqrt{3} \).

The ships are on opposite sides of the light house, and they are 230\(\sqrt{3}\) m apart. The distance between the ships C and D is the sum of their distances from the base B, i.e., \( CD = BC + BD = x_2 + x_1 \).

We are given \( x_1 + x_2 = 230\sqrt{3} \).

Substitute the expressions for \(x_1\) and \(x_2\) in terms of H:

\( \frac{H}{\sqrt{3}} + H\sqrt{3} = 230\sqrt{3} \)

To solve for H, we can find a common denominator or multiply the entire equation by \(\sqrt{3}\):

\( \sqrt{3} \times \left( \frac{H}{\sqrt{3}} + H\sqrt{3} \right) = \sqrt{3} \times \left( 230\sqrt{3} \right) \)

\( H + H(\sqrt{3})(\sqrt{3}) = 230(\sqrt{3})(\sqrt{3}) \)

\( H + 3H = 230 \times 3 \)

\( 4H = 690 \)

Now, divide by 4 to find H:

\( H = \frac{690}{4} \)

\( H = 172.5 \)

The height of the light house is 172.5 meters.

Step-by-Step Solution

1. Identify the given information: Angles of depression (30° and 60°), distance between ships (230\(\sqrt{3}\) m), ships are on opposite sides and collinear with the light house.
2. Understand that angles of elevation from ships to the light house top are equal to the angles of depression.
3. Draw a diagram representing the light house as a vertical line and the ships on a horizontal line on opposite sides of the base.
4. Set up trigonometric equations (using tangent) for the two right-angled triangles formed by the light house, the base, and each ship.
5. Let H be the height of the light house, \(x_1\) and \(x_2\) be the distances of the ships from the base.
6. From \(\tan(60^\circ)\), express \(x_1\) in terms of H: \(x_1 = \frac{H}{\sqrt{3}}\).
7. From \(\tan(30^\circ)\), express \(x_2\) in terms of H: \(x_2 = H\sqrt{3}\).
8. Use the given total distance between ships: \(x_1 + x_2 = 230\sqrt{3}\).
9. Substitute the expressions for \(x_1\) and \(x_2\) into the total distance equation.
10. Solve the resulting equation for H: \( \frac{H}{\sqrt{3}} + H\sqrt{3} = 230\sqrt{3} \implies 4H = 690 \implies H = 172.5 \).
11. State the final answer: The height of the light house is 172.5 m.

Final Calculation

\( \frac{H}{\sqrt{3}} + H\sqrt{3} = 230\sqrt{3} \)

Multiply by \( \sqrt{3} \):

\( H + 3H = 230 \times 3 \)

\( 4H = 690 \)

\( H = \frac{690}{4} = 172.5 \)

The height of the light house is 172.5 m.

Revision Table: Trigonometry and Heights

Additional Information: Angles of Elevation and Depression

Angles of elevation and depression are crucial concepts in trigonometry, often used in problems involving heights and distances. They are always measured between a horizontal line and the line of sight.

• Angle of Elevation: The angle formed by the line of sight and the horizontal line when the object is above the horizontal line (looking upwards).
• Angle of Depression: The angle formed by the line of sight and the horizontal line when the object is below the horizontal line (looking downwards).

In geometry, if a horizontal line is considered parallel to the ground, the angle of depression from point A to point B is equal to the angle of elevation from point B to point A. This is because they are alternate interior angles formed by a transversal line (the line of sight) intersecting two parallel lines (the horizontal line at A and the ground level at B).