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
Understanding the Aeroplane Height Problem
This problem involves using trigonometry to find the height of an aeroplane above a horizontal road. We are given the angles of depression from the aeroplane to two consecutive kilometre-stones located on opposite sides of the point directly below the aeroplane.
Key Concepts: Angle of Depression and Elevation
The angle of depression is the angle between the horizontal line from the observer's eye to an object below the horizontal line. The angle of elevation is the angle between the horizontal line from the observer's eye to an object above the horizontal line.
- When an observer looks down at an object, the angle of depression is formed.
- When an observer looks up at an object, the angle of elevation is formed.
- Importantly, if a horizontal line is 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 (due to alternate interior angles).
Setting up the Geometry
Let's visualize the scenario:
- Let P be the position of the aeroplane.
- Let O be the point on the road directly below the aeroplane (so PO is the height).
- Let A and B be the positions of the two consecutive kilometre-stones on the road.
- The stones A and B are on opposite sides of O.
- The distance between the stones AB is 1 kilometre.
- The height of the aeroplane above the road is PO = h (which we need to find).
The angle of depression from P to A is given as \(\alpha\). This means the angle of elevation from A to P is also \(\alpha\).
The angle of depression from P to B is given as \(\beta\). This means the angle of elevation from B to P is also \(\beta\).
We have two right-angled triangles: \(\triangle POA\) and \(\triangle POB\). Both triangles share the common side PO, which is the height h.
Using Trigonometry to Find Height
In the right-angled triangle \(\triangle POA\):
- The side opposite to the angle \(\alpha\) (at A) is PO = h.
- The side adjacent to the angle \(\alpha\) is OA (the distance from O to stone A).
Using the tangent function (\(\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}\)):
\(\tan \alpha = \frac{PO}{OA} = \frac{h}{OA}\)
From this, we can find the distance OA:
\(OA = \frac{h}{\tan \alpha} = h \cot \alpha\) (Since \(\cot \alpha = \frac{1}{\tan \alpha}\))
Now, in the right-angled triangle \(\triangle POB\):
- The side opposite to the angle \(\beta\) (at B) is PO = h.
- The side adjacent to the angle \(\beta\) is OB (the distance from O to stone B).
Using the tangent function:
\(\tan \beta = \frac{PO}{OB} = \frac{h}{OB}\)
From this, we can find the distance OB:
\(OB = \frac{h}{\tan \beta} = h \cot \beta\) (Since \(\cot \beta = \frac{1}{\tan \beta}\))
Calculating the Total Distance
Since the two kilometre-stones A and B are on opposite sides of the point O directly below the aeroplane, the total distance between them is the sum of OA and OB.
Distance AB = OA + OB
We know that the stones are consecutive kilometre-stones, so the distance AB is 1 km.
\(1 = OA + OB\)
Substitute the expressions for OA and OB that we found using trigonometry:
\(1 = h \cot \alpha + h \cot \beta\)
Solving for the Aeroplane's Height
Now we need to solve this equation for h:
\(1 = h (\cot \alpha + \cot \beta)\)
Divide both sides by \((\cot \alpha + \cot \beta)\) to isolate h:
\(h = \frac{1}{\cot \alpha + \cot \beta}\)
This gives the height in terms of cotangents. The options are given in terms of tangents. Let's convert the expression using the identity \(\cot x = \frac{1}{\tan x}\).
\(h = \frac{1}{\frac{1}{\tan \alpha} + \frac{1}{\tan \beta}}\)
To simplify the denominator, find a common denominator:
The common denominator for \(\frac{1}{\tan \alpha} + \frac{1}{\tan \beta}\) is \(\tan \alpha \tan \beta\).
\(\frac{1}{\tan \alpha} + \frac{1}{\tan \beta} = \frac{1 \times \tan \beta}{\tan \alpha \times \tan \beta} + \frac{1 \times \tan \alpha}{\tan \beta \times \tan \alpha} = \frac{\tan \beta + \tan \alpha}{\tan \alpha \tan \beta}\)
Substitute this back into the expression for h:
\(h = \frac{1}{\frac{\tan \alpha + \tan \beta}{\tan \alpha \tan \beta}}\)
When dividing by a fraction, we multiply by its reciprocal:
\(h = 1 \times \frac{\tan \alpha \tan \beta}{\tan \alpha + \tan \beta}\)
\(h = \frac{\tan \alpha \tan \beta}{\tan \alpha + \tan \beta}\)
Conclusion on Aeroplane Height
The height of the aeroplane above the road is \(\frac{\tan \alpha \tan \beta}{\tan \alpha + \tan \beta}\). This matches one of the given options.
| Variable/Concept | Description | Mathematical Representation |
|---|---|---|
| h | Height of the aeroplane | PO |
| \(\alpha\) | Angle of depression to stone A (or elevation from A) | |
| \(\beta\) | Angle of depression to stone B (or elevation from B) | |
| OA | Distance from O to stone A | \(h \cot \alpha\) |
| OB | Distance from O to stone B | \(h \cot \beta\) |
| AB | Distance between stones A and B | 1 km |
Revision Table: Height Calculation Summary
| Step | Action | Result |
|---|---|---|
| 1 | Relate angles of depression to angles of elevation. | Angle A = \(\alpha\), Angle B = \(\beta\) |
| 2 | Use trigonometry (tan) in \(\triangle POA\). | \(OA = h \cot \alpha\) |
| 3 | Use trigonometry (tan) in \(\triangle POB\). | \(OB = h \cot \beta\) |
| 4 | Use the fact that stones are on opposite sides and 1 km apart. | \(OA + OB = 1\) |
| 5 | Substitute OA and OB expressions. | \(h \cot \alpha + h \cot \beta = 1\) |
| 6 | Factor out h and solve. | \(h = \frac{1}{\cot \alpha + \cot \beta}\) |
| 7 | Convert to tangent form. | \(h = \frac{\tan \alpha \tan \beta}{\tan \alpha + \tan \beta}\) |
Additional Information: Applications of Trigonometry
Trigonometry, particularly the concepts of angles of elevation and depression, is widely used in various real-world applications:
- Navigation: Pilots and sailors use these angles for determining positions and distances.
- Surveying: Surveyors use angles to measure heights of buildings, trees, or mountains and distances between points.
- Architecture and Engineering: Calculating heights, angles, and slopes in construction projects.
- Astronomy: Measuring distances to celestial bodies.
- Physics: Analyzing projectile motion and forces.
Understanding how to apply trigonometric ratios (sine, cosine, tangent, cotangent, secant, cosecant) based on given angles and side lengths is fundamental to solving such practical problems.
Similar Questions
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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 ?
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