Consider the following for the next two (02) items that follow : The angle of elevation of a cloud at C from a point (P), H metres above the surface of a lake is 30°. The height of the cloud from the surface of the lake is 2H metres. Let θ be the angle of depression of the reflection of the cloud in the lake from the point P.
What is the value of θ ?
Solving the Angle of Depression Problem with Reflection
This problem involves understanding angles of elevation and depression, along with the concept of reflection in a surface like a lake. We are given the height of an observation point P above the lake, the height of a cloud C above the lake, and the angle of elevation to the cloud. We need to find the angle of depression to the reflection of the cloud in the lake.
Understanding the Setup and Key Concepts
- The point P is H metres above the lake surface.
- The cloud C is 2H metres above the lake surface.
- The angle of elevation of the cloud C from P is 30°. This is the angle between the horizontal line from P and the line of sight PC.
- The reflection of the cloud in the lake, let's call it C', appears to be as far below the lake surface as the cloud is above it. So, C' is 2H metres below the lake surface.
- The angle of depression of the reflection C' from P is θ. This is the angle between the horizontal line from P and the line of sight PC'. We need to find θ.
Let's visualize this. Draw a horizontal line representing the lake surface. Point P is H above it. Cloud C is 2H above it. Reflection C' is 2H below it.
Draw a horizontal line through P. The vertical distance from P to the lake surface is H. The vertical distance from the horizontal line through P down to the lake surface is H.
Setting up Triangles for Calculation
Consider a vertical line passing through the cloud C and its reflection C'. Let A be the point on this vertical line that is at the same height as P (i.e., on the horizontal line through P). Let O be the point where this vertical line intersects the lake surface.
The height of C from the lake surface is CO = 2H.
The depth of C' from the lake surface is C'O = 2H.
The height of P from the lake surface is PO' = H, where O' is the point on the lake surface directly below P. The horizontal line through P is at height H above the lake.
The vertical distance from the horizontal line through P up to the cloud C is CA. This distance is CO - AO = 2H - H = H.
The vertical distance from the horizontal line through P down to the reflection C' is AC'. This distance is AO + OC' = H + 2H = 3H.
Let the horizontal distance from P to the vertical line containing C and C' be x (this is the distance PA). The horizontal distance is the same for both triangles (P to C and P to C').
For the angle of elevation of the cloud C from P (30°):
In the right-angled triangle PAC, the angle of elevation is 30°. The opposite side is CA = H, and the adjacent side is PA = x.
Using trigonometry:
\(\tan(30^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{CA}{PA} = \frac{H}{x}\)
From this, we can find x:
\(x = \frac{H}{\tan(30^\circ)}\)
We know that \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\).
So, \(x = \frac{H}{1/\sqrt{3}} = H\sqrt{3}\)
For the angle of depression of the reflection C' from P (θ):
In the right-angled triangle PAC', the angle of depression is θ. The vertical distance from the horizontal line through P down to C' is AC' = 3H. The horizontal distance is PA = x.
Using trigonometry:
\(\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AC'}{PA} = \frac{3H}{x}\)
Now, substitute the value of x we found:
\(\tan(\theta) = \frac{3H}{H\sqrt{3}}\)
\(\tan(\theta) = \frac{3}{\sqrt{3}}\)
To simplify \(\frac{3}{\sqrt{3}}\), we can multiply the numerator and denominator by \(\sqrt{3}\):
\(\tan(\theta) = \frac{3}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{3}}{3} = \sqrt{3}\)
So, we have \(\tan(\theta) = \sqrt{3}\).
We need to find the angle θ whose tangent is \(\sqrt{3}\). This is a standard trigonometric value.
\(\theta = \arctan(\sqrt{3})\)
The angle is 60°.
Therefore, the value of θ is 60°.
Let's summarize the key distances from the horizontal line passing through P:
- Height of P above lake: H
- Height of cloud C above lake: 2H
- Depth of reflection C' below lake: 2H
- Vertical distance from horizontal through P to C: 2H - H = H
- Vertical distance from horizontal through P to C': H + 2H = 3H
If the horizontal distance from P to the cloud/reflection's vertical line is x:
- Angle of elevation (to C): \(\tan(30^\circ) = \frac{H}{x} \implies x = \frac{H}{\tan(30^\circ)}\)
- Angle of depression (to C'): \(\tan(\theta) = \frac{3H}{x}\)
Substituting x:
\(\tan(\theta) = \frac{3H}{H/\tan(30^\circ)} = 3H \times \frac{\tan(30^\circ)}{H} = 3 \times \tan(30^\circ)\)
\(\tan(\theta) = 3 \times \frac{1}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}\)
Since \(\tan(60^\circ) = \sqrt{3}\), \(\theta = 60^\circ\).
Revision Table: Geometry Concepts
| Concept | Definition | Application in Problem |
|---|---|---|
| Angle of Elevation | Angle between the horizontal line of sight and the line of sight to an object above the horizontal. | Angle to the cloud from point P (30°). |
| Angle of Depression | Angle between the horizontal line of sight and the line of sight to an object below the horizontal. | Angle to the reflection of the cloud from point P (θ). |
| Reflection in a Lake | Image appears as far below the surface as the object is above it. | Cloud at 2H above surface means reflection is at 2H below surface. |
| Trigonometric Ratios | Relationships between angles and side lengths in right-angled triangles (e.g., tan = Opposite/Adjacent). | Used to relate angles 30° and θ to vertical and horizontal distances. |
Additional Information: Angles and Heights in Trigonometry
Problems involving heights and distances often use angles of elevation and depression. It is crucial to draw a diagram and identify the right-angled triangles. The horizontal line from the observation point is key to defining these angles.
When dealing with reflections in water, the surface acts like a mirror. The image is virtual and located symmetrically opposite to the object with respect to the surface. This means the perpendicular distance from the object to the surface is equal to the perpendicular distance from the image (reflection) to the surface.
Standard trigonometric values for common angles like 0°, 30°, 45°, 60°, and 90° are often used in these problems. Remembering that \(\tan(30^\circ) = 1/\sqrt{3}\) and \(\tan(60^\circ) = \sqrt{3}\) is helpful.
The calculation involved finding a common horizontal distance using the known angle of elevation and height difference, then using this distance with the height difference to the reflection to find the angle of depression. The height difference to the reflection from the horizontal line through P is the height of P above the lake plus the depth of the reflection below the lake.
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Important Questions from Heights and Distances
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