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Question

The angle of elevation of the top of a hill at the foot of the tower is 60° and the angle of elevation of the top of the tower from the foot of the hill is 30°. If the tower is 50 m high, what is the height of the hill?

This question was previously asked in
RRB ALP 2018 CBT 2 Fitter Question Paper (21-Jan-2019) (Shift 3)
The correct answer is

150 m

Understanding the Problem: Height of Hill Calculation

This problem involves trigonometry, specifically using angles of elevation to find unknown heights. We have a tower and a hill, and we are given the heights of one (the tower) and angles of elevation between the top of one structure and the foot of the other.

Defining Angles of Elevation

The angle of elevation is the angle between the horizontal line from an observer's eye to an object, when the object is above the horizontal line. In this case, the horizontal line is the ground level between the tower and the hill.

  • The angle of elevation of the top of the hill at the foot of the tower is 60°. This means if you stand at the base of the tower and look up at the top of the hill, the angle your line of sight makes with the horizontal ground is 60°.
  • The angle of elevation of the top of the tower from the foot of the hill is 30°. This means if you stand at the base of the hill and look up at the top of the tower, the angle your line of sight makes with the horizontal ground is 30°.

We are given the height of the tower is 50 m. We need to find the height of the hill.

Setting up the Triangles

Let's imagine the situation forms two right-angled triangles. Let:

  • \(H\) be the height of the hill.
  • \(T\) be the height of the tower, so \(T = 50\) m.
  • \(D\) be the horizontal distance between the foot of the tower and the foot of the hill.

From the problem description, we can form two right-angled triangles:

  1. Triangle formed by the tower's height, the horizontal distance, and the line of sight from the foot of the hill to the top of the tower. In this triangle, the angle of elevation is 30°, the opposite side is \(T = 50\) m, and the adjacent side is \(D\).
  2. Triangle formed by the hill's height, the horizontal distance, and the line of sight from the foot of the tower to the top of the hill. In this triangle, the angle of elevation is 60°, the opposite side is \(H\), and the adjacent side is \(D\).

Using Trigonometry to Solve

We can use the tangent function, which relates the opposite side and the adjacent side in a right-angled triangle: \(\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}\).

Step 1: Find the Horizontal Distance (D)

From Triangle 1 (using the tower):

\(\tan(30^\circ) = \frac{\text{Height of Tower}}{\text{Distance Between Bases}}\)

\(\tan(30^\circ) = \frac{50}{D}\)

We know that \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\). So,

\(\frac{1}{\sqrt{3}} = \frac{50}{D}\)

Solving for \(D\):

\(D = 50 \times \sqrt{3}\) m

Step 2: Find the Height of the Hill (H)

From Triangle 2 (using the hill):

\(\tan(60^\circ) = \frac{\text{Height of Hill}}{\text{Distance Between Bases}}\)

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

We know that \(\tan(60^\circ) = \sqrt{3}\). So,

\(\sqrt{3} = \frac{H}{D}\)

Now, substitute the value of \(D\) we found in Step 1:

\(\sqrt{3} = \frac{H}{50\sqrt{3}}\)

Solving for \(H\):

\(H = \sqrt{3} \times (50\sqrt{3})\)

\(H = 50 \times (\sqrt{3} \times \sqrt{3})\)

\(H = 50 \times 3\)

\(H = 150\) m

So, the height of the hill is 150 m.

Summary of Calculation

We used the given angles of elevation and the height of the tower to first find the horizontal distance between the two structures, and then used that distance along with the other angle of elevation to find the height of the hill.

Revision Table: Trigonometry Basics

Trigonometric Ratio Definition Common Values
Sine (\(\sin \theta\)) Opposite / Hypotenuse \(\sin 30^\circ = 1/2\), \(\sin 60^\circ = \sqrt{3}/2\)
Cosine (\(\cos \theta\)) Adjacent / Hypotenuse \(\cos 30^\circ = \sqrt{3}/2\), \(\cos 60^\circ = 1/2\)
Tangent (\(\tan \theta\)) Opposite / Adjacent \(\tan 30^\circ = 1/\sqrt{3}\), \(\tan 60^\circ = \sqrt{3}\)

Additional Information: Applications of Elevation Angles

Angles of elevation and depression are widely used in various fields:

  • Surveying: To determine the height of buildings, trees, or geographical features without physically measuring them.
  • Navigation: In aerial and maritime navigation to determine positions and distances.
  • Astronomy: To measure the altitude of celestial bodies.
  • Engineering: In construction and design, such as planning ramps or bridges.

These angles are crucial tools for indirect measurement using the principles of trigonometry.

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Similar Questions

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  2. A ladder 13 m long reaches a window which is 12 m above the ground on side of a street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 5 m high, then the width of the street is:

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Important Questions from Heights and Distances

  1. Mohit is standing at some distance from a 60 meters tall building. Mohit is 1.8 meters tall. When Mohit walks towards the building, then the angle of elevation from his head becomes 60° from 45°. How much distance (in metres) Mohit covered towards the building?

  2. A peacock sitting at the top of a 3 meter high pole saw a snake approaching towards pole at a distance three times of the height of the pole. Then it jumping from pole will catch the snake at what distance from the pole if both are running with same speed ?

  3. The foot of a ladder 25 m long is 7 m from the base of the building. If the top of the ladder slips by 4 m, then by how much distance will the foot of the ladder slide?

  4. Two hotels stand 25 m apart. One of them is 70 m high and the angle of depression of the top of other as observed from the top of this hotel is 45°. Height of the other hotel is:

  5. A 7 m 20 cm pole casts a shadow of length 8 m 30 cm. Find the height of a tree that casts a shadow of length 6 m 64 cm, under similar conditions.

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