The angle of elevation of the top of a tower of height x metre from a point on the ground is found to be 60°. By going y metre away from that point, it becomes 30°. Which one of the following relations is correct?
Solving the Angle of Elevation Problem
This problem involves the concept of angles of elevation and trigonometric ratios. We are given the height of a tower and the angles of elevation from two different points on the ground. We need to find the relationship between the tower's height and the distance moved away from the first observation point.
Setting up the Geometry
Let's visualize the situation. We have a tower standing vertically on the ground. Let the height of the tower be \(h = x\) metres. Let the first observation point on the ground be A and the base of the tower be B. The angle of elevation from A to the top of the tower (say T) is 60°.
So, in the right-angled triangle ABT, the angle ∠TAB = 60°. Let the distance AB be \(d\) metres.
The observer then moves \(y\) metres away from point A to a new point C, which is further away from the base of the tower. The distance BC will be \(d + y\) metres. The angle of elevation from point C to the top of the tower T is 30°. So, in the right-angled triangle CBT, the angle ∠TCB = 30°.
Using Trigonometric Ratios
In a right-angled triangle, the tangent of an angle is the ratio of the opposite side to the adjacent side.
Triangle ABT (Angle 60°)
The opposite side to the 60° angle is the height of the tower, \(x\). The adjacent side is the distance from the base to point A, \(d\).
Using the tangent ratio:
\(\tan(60^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{x}{d}\)
We know that \(\tan(60^\circ) = \sqrt{3}\). So,
\(\sqrt{3} = \frac{x}{d}\)
From this equation, we can express \(d\) in terms of \(x\):
\(d = \frac{x}{\sqrt{3}}\) (Equation 1)
Triangle CBT (Angle 30°)
The opposite side to the 30° angle is still the height of the tower, \(x\). The adjacent side is the distance from the base to point C, which is \(d + y\).
Using the tangent ratio:
\(\tan(30^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{x}{d + y}\)
We know that \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\). So,
\(\frac{1}{\sqrt{3}} = \frac{x}{d + y}\)
Rearranging this equation, we get:
\(d + y = x\sqrt{3}\) (Equation 2)
Finding the Relationship between x and y
Now we have two equations. We can substitute the value of \(d\) from Equation 1 into Equation 2.
Substitute \(d = \frac{x}{\sqrt{3}}\) into \(d + y = x\sqrt{3}\):
\(\frac{x}{\sqrt{3}} + y = x\sqrt{3}\)
We want to find the relationship between \(x\) and \(y\), so let's isolate \(y\):
\(y = x\sqrt{3} - \frac{x}{\sqrt{3}}\)
To combine the terms on the right side, find a common denominator (\(\sqrt{3}\)):
\(y = \frac{x\sqrt{3} \cdot \sqrt{3}}{\sqrt{3}} - \frac{x}{\sqrt{3}}\)
\(y = \frac{3x}{\sqrt{3}} - \frac{x}{\sqrt{3}}\)
\(y = \frac{3x - x}{\sqrt{3}}\)
\(y = \frac{2x}{\sqrt{3}}\)
To get rid of the fraction, multiply both sides by \(\sqrt{3}\):
\(y\sqrt{3} = 2x\)
This can also be written as:
\(2x = y\sqrt{3}\) or \(2x = \sqrt{3}y\)
Comparing with Options
Let's compare our derived relationship \(2x = \sqrt{3}y\) with the given options:
- Option 1: \(x = y\) (Incorrect)
- Option 2: \(2x = 3y\) (Incorrect)
- Option 3: \(2x = \sqrt{3}y\) (Correct)
- Option 4: \(2y = \sqrt{3}x\) (Incorrect)
Our result matches Option 3.
Step-by-Step Summary
- Identify the two right-angled triangles formed by the tower, the ground, and the lines of sight.
- Use the tangent trigonometric ratio in the first triangle (angle 60°) to relate the height \(x\) and the initial distance \(d\).
- Use the tangent trigonometric ratio in the second triangle (angle 30°) to relate the height \(x\) and the new distance \(d+y\).
- Solve the two equations simultaneously to eliminate the unknown distance \(d\).
- Simplify the resulting equation to find the relationship between \(x\) and \(y\).
| Angle of Elevation | Triangle | Base Length | Trigonometric Equation |
|---|---|---|---|
| 60° | ABT | \(d\) | \(\tan(60^\circ) = \frac{x}{d} \Rightarrow \sqrt{3} = \frac{x}{d}\) |
| 30° | CBT | \(d+y\) | \(\tan(30^\circ) = \frac{x}{d+y} \Rightarrow \frac{1}{\sqrt{3}} = \frac{x}{d+y}\) |
Revision Table: Key Concepts
| Concept | Description | Relevance to Problem |
|---|---|---|
| Angle of Elevation | The angle between the horizontal line of sight and the line of sight upwards to an object. | Given as 60° and 30° in the problem. |
| Right-Angled Triangle | A triangle with one angle equal to 90°. | Formed by the tower, the ground, and the line of sight. |
| Trigonometric Ratios | Ratios of sides in a right-angled triangle (sine, cosine, tangent). | Tangent ratio (\(\tan\)) is used as it relates the opposite side (height) and adjacent side (distance). |
| Tangent (\(\tan\)) | Opposite side / Adjacent side | \(\tan(60^\circ) = \sqrt{3}\), \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\) |
Additional Information: Angles and Trigonometry
Angles of elevation and depression are commonly used in trigonometry problems involving heights and distances. The angle of elevation is always measured upwards from a horizontal line, while the angle of depression is measured downwards from a horizontal line.
The basic trigonometric ratios (sine, cosine, and tangent) are defined based on the sides of a right-angled triangle relative to a specific acute angle:
- Sine (\(\sin\)): Opposite side / Hypotenuse
- Cosine (\(\cos\)): Adjacent side / Hypotenuse
- Tangent (\(\tan\)): Opposite side / Adjacent side
These ratios have specific values for common angles like 0°, 30°, 45°, 60°, and 90°. Remembering these values is crucial for solving many trigonometry problems without a calculator.
In this problem, we used the tangent ratio because we were relating the vertical height (opposite side) to the horizontal distance (adjacent side). When dealing with angles of elevation or depression from a point on the ground to the top of a vertical object, the tangent ratio is frequently the most useful.
Similar Questions
A person on the top of a vertical tower observes a car moving at a uniform speed coming directly towards it. If it takes 6 minutes for the angle of depression to change from 30° to 45°, and further t minutes to reach the tower, which one of the following is correct ?
From an aeroplane vertically over a straight horizontal road, the angle of depression of two consecutive kilometre-stones on the opposite sides of the aeroplane are observed to be α and β. The height of the aeroplane above the road is
Two observers are stationed due north of a tower (of height x meter) at a distance y meter from each other. The angles of elevation of the tower observed by them are 30° and 45° respectively. Then x/y is equal to
What is the value of θ ?
- What is PC equal to ?
A plane is going in circles around an airport. The plane takes 3 minutes to complete one round. The angle of elevation of the plane from a point P on the ground at time t seconds is equal to that at time (t + 30) seconds. At time (t + x) seconds, the plane flies vertically above the point P. What is x equal to?
From a height of h units, a man observes the angle of elevation as α and angle of depression as β of the top and bottom respectively of a tower of height H (> 4h). To what further height should be climb so that the values of angle of elevation and angle of depression get interchanged for the top and bottom of the tower?
ABC is a right angles triangle with base BC and height AB. The hypotenuse AC is four times the length of the perpendicular drawn to it from opposite vertex. What is tan C equal to?
A pole stands vertically inside a triangular park ABC. If the angle of elevation of the top of the pole from each corner of the park is same, then in ΔABC, the foot of the pole is at the:
From an aeroplane flying above a river at an altitude of 1200 m, it is observed that the angles of depression of opposite points on the two banks of a river are 30° and θ. If the width of the river is 3000 m, then which one of the following is correct?
Important Questions from Heights and Distances
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 ?
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?
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:
If the angles of elevation of a balloon from two consecutive kilometer-stones along a straight road are 30° and 60° respectively, then the height of the balloon above the ground will be:
The angle of elevation of the top of a tower from a point 20 m away from its base is 45 °. What is the height of the tower?
