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 PC equal to ?
Understanding the Cloud Elevation Problem
This question asks us to find the distance between a point P, located H metres above a lake, and a cloud C, which is 2H metres above the lake surface. We are given that the angle of elevation of the cloud from point P is 30°. We will use basic trigonometry to solve this geometry problem.
Setting up the Geometry
Let's visualize the scenario:
- Let L be the surface of the lake.
- Point P is H metres above L.
- Cloud C is 2H metres above L.
- Draw a horizontal line from P that meets the vertical line from C to the lake surface at point Q. This forms a right-angled triangle PQC, where the angle at Q is 90°.
The height of P above the lake is H. The height of C above the lake is 2H.
Since PQ is horizontal and P is H metres above the lake, Q is also H metres above the lake surface. The vertical distance from C to Q is the difference between the height of C and the height of Q above the lake surface.
Vertical distance CQ = Height of C above lake - Height of Q above lake
Vertical distance CQ = \(2H - H = H\) metres.
In the right-angled triangle PQC, we know:
- The angle of elevation of the cloud C from P is \(\angle CPQ = 30^\circ\).
- The side opposite to this angle is CQ, which we found to be H metres.
- We need to find the distance PC, which is the hypotenuse of the triangle PQC.
Using Trigonometry to Find PC
In a right-angled triangle, the sine of an angle is defined as the ratio of the length of the opposite side to the length of the hypotenuse. For triangle PQC and angle \(\angle CPQ\):
\(\sin(\angle CPQ) = \frac{\text{Opposite side}}{\text{Hypotenuse}}\)
\(\sin(30^\circ) = \frac{CQ}{PC}\)
We know CQ = H and \(\sin(30^\circ) = \frac{1}{2}\). Substituting these values:
\(\frac{1}{2} = \frac{H}{PC}\)
Now, we can solve for PC:
\(PC \times \frac{1}{2} = H\)
\(PC = 2 \times H\)
\(PC = 2H\)
So, the distance PC is 2H metres.
Comparing with Options
Let's look at the given options:
- H metres
- \(\sqrt{2}H\) metres
- \(\sqrt{3}H\) metres
- 2H metres
Our calculated value for PC is 2H metres, which matches the fourth option.
Summary of Calculation
| Parameter | Value/Calculation |
|---|---|
| Height of P above lake | H |
| Height of C above lake | 2H |
| Vertical distance CQ | \(2H - H = H\) |
| Angle of elevation \(\angle CPQ\) | 30° |
| Trigonometric relation | \(\sin(30^\circ) = \frac{CQ}{PC}\) |
| Solving for PC | \(\frac{1}{2} = \frac{H}{PC} \implies PC = 2H\) |
Conclusion
Based on the angle of elevation and the relative heights of point P and cloud C above the lake, the distance PC is found to be 2H metres.
Trigonometry Revision Table
| Angle (\(\theta\)) | \(\sin(\theta)\) | \(\cos(\theta)\) | \(\tan(\theta)\) |
|---|---|---|---|
| 0° | 0 | 1 | 0 |
| 30° | \(\frac{1}{2}\) | \(\frac{\sqrt{3}}{2}\) | \(\frac{1}{\sqrt{3}}\) |
| 45° | \(\frac{1}{\sqrt{2}}\) | \(\frac{1}{\sqrt{2}}\) | 1 |
| 60° | \(\frac{\sqrt{3}}{2}\) | \(\frac{1}{2}\) | \(\sqrt{3}\) |
| 90° | 1 | 0 | Undefined |
Additional Information on Angles of Elevation and Depression
The concept of angles of elevation and depression is fundamental in trigonometry, especially in height and distance problems.
- Angle of Elevation: This is the angle between the horizontal line from the observer's eye and the line of sight to an object above the horizontal line. Imagine looking upwards at something; the angle your gaze makes with the horizontal is the angle of elevation.
- Angle of Depression: This is the angle between the horizontal line from the observer's eye and the line of sight to an object below the horizontal line. Imagine looking downwards at something; the angle your gaze makes with the horizontal is the angle of depression.
- These angles are measured from the horizontal. In problems involving reflections in lakes or water bodies, the reflection is considered to be as far below the surface as the object is above, along the same vertical line. The reflection acts like an image formed by a mirror.
- For this specific problem, the information about the angle of depression of the reflection is part of a subsequent question and wasn't needed to calculate the distance PC using the angle of elevation to the cloud itself.
Similar Questions
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Important Questions from Heights and Distances
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