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 ?
8 < t < 8.3
Understanding the Angle of Depression Problem
This question involves trigonometry, specifically the concept of angles of depression. The scenario describes a person on top of a vertical tower observing a car moving towards the tower at a uniform speed. We are given information about how the angle of depression changes over time and need to determine the time it takes for the car to reach the tower from a certain point.
The angle of depression is the angle between the horizontal line from the observer's eye and the line of sight to an object below the horizontal line. When a car is observed from a tower, the angle of depression to the car is numerically equal to the angle of elevation of the tower from the car's position (assuming the ground is horizontal and the tower is vertical).
Setting up the Problem with Trigonometry
Let's visualize the situation:
- Let \(h\) be the height of the vertical tower.
- Let the initial position of the car be A, the position after 6 minutes be B, and the base of the tower be C. The observer is at the top of the tower, say at point D.
- The car moves from A to B in 6 minutes and then from B to C in \(t\) minutes.
- The angles of depression to the car at points A and B are 30° and 45°, respectively. This means the angles of elevation from A and B to the top of the tower D are also 30° and 45°.
- The car moves at a uniform speed, let's call it \(v\).
We have two right-angled triangles formed: \(\triangle DBC\) and \(\triangle DAC\).
In \(\triangle DBC\), the angle at B is 45° and the height is DC = \(h\). The distance BC is the distance of the car from the tower base when the angle of depression is 45°.
Using trigonometry (tangent function):
\(\tan(\text{angle of elevation}) = \frac{\text{Opposite side}}{\text{Adjacent side}}\)
For \(\triangle DBC\):
\(\tan(45^\circ) = \frac{DC}{BC}\)
\(1 = \frac{h}{BC}\)
So, \(BC = h\).
In \(\triangle DAC\), the angle at A is 30° and the height is DC = \(h\). The distance AC is the distance of the car from the tower base when the angle of depression is 30°.
For \(\triangle DAC\):
\(\tan(30^\circ) = \frac{DC}{AC}\)
\(\frac{1}{\sqrt{3}} = \frac{h}{AC}\)
So, \(AC = h\sqrt{3}\).
Calculating Distances and Speed
The distance the car traveled from A to B is \(AB = AC - BC\).
\(AB = h\sqrt{3} - h = h(\sqrt{3} - 1)\).
The car travels this distance AB in 6 minutes. Since the speed is uniform, the distance is also given by speed \(\times\) time.
\(AB = v \times 6\)
So, \(h(\sqrt{3} - 1) = 6v\).
The car then travels from B to C (the base of the tower) in \(t\) minutes. The distance BC is \(h\), as calculated earlier.
Using the uniform speed again:
\(BC = v \times t\)
\(h = vt\).
Finding the Time 't'
We have two equations relating \(h\), \(v\), and \(t\):
- \(h(\sqrt{3} - 1) = 6v\)
- \(h = vt\)
From equation (2), we can express \(v\) as \(\frac{h}{t}\). Substitute this into equation (1):
\(h(\sqrt{3} - 1) = 6 \left(\frac{h}{t}\right)\)
Assuming the tower has a height (\(h \ne 0\)), we can divide both sides by \(h\):
\(\sqrt{3} - 1 = \frac{6}{t}\)
Now, solve for \(t\):
\(t = \frac{6}{\sqrt{3} - 1}\)
Rationalizing and Approximating the Value of 't'
To get a numerical value, we should rationalize the denominator and use the approximate value of \(\sqrt{3}\).
Rationalize the denominator:
\(t = \frac{6}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1}\)
\(t = \frac{6(\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2}\)
\(t = \frac{6(\sqrt{3} + 1)}{3 - 1}\)
\(t = \frac{6(\sqrt{3} + 1)}{2}\)
\(t = 3(\sqrt{3} + 1)\)
Using the approximate value \(\sqrt{3} \approx 1.732\):
\(t \approx 3(1.732 + 1)\)
\(t \approx 3(2.732)\)
\(t \approx 8.196\)
Comparing with the Options
The calculated time \(t\) is approximately 8.196 minutes. We need to find which range this value falls into from the given options:
- 7.7 < t < 8 (8.196 is not in this range)
- 8 < t < 8.3 (8.196 is in this range)
- 8.3 < t < 8.6 (8.196 is not in this range)
- 8.6 < t < 8.9 (8.196 is not in this range)
The value \(t \approx 8.196\) minutes falls within the range \(8 < t < 8.3\).
Summary of Steps
- Identify the angles of elevation corresponding to the angles of depression.
- Use the tangent function in the right-angled triangles formed by the tower and the car's positions.
- Express the distances traveled by the car in terms of the height of the tower \(h\).
- Relate the distances traveled to the uniform speed \(v\) and the given times (6 minutes and \(t\) minutes).
- Solve the system of equations to find the value of \(t\).
- Calculate the approximate numerical value of \(t\) and compare it with the given options.
| Angle of Depression | Angle of Elevation | Distance from Tower Base |
|---|---|---|
| 30° | 30° | \(AC = h\sqrt{3}\) |
| 45° | 45° | \(BC = h\) |
Distance covered in 6 minutes = \(AB = AC - BC = h\sqrt{3} - h = h(\sqrt{3} - 1)\). This distance is also \(6v\). So, \(h(\sqrt{3}-1) = 6v\).
Distance covered in \(t\) minutes (from 45° point to tower base) = \(BC = h\). This distance is also \(vt\). So, \(h = vt\).
From \(h=vt\), \(v = h/t\). Substitute into the first equation:
\(h(\sqrt{3}-1) = 6(h/t)\)
Assuming \(h \ne 0\), divide by \(h\):
\(\sqrt{3}-1 = 6/t\)
\(t = \frac{6}{\sqrt{3}-1} = \frac{6(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{6(\sqrt{3}+1)}{3-1} = \frac{6(\sqrt{3}+1)}{2} = 3(\sqrt{3}+1)\).
Using \(\sqrt{3} \approx 1.732\), \(t \approx 3(1.732+1) = 3(2.732) = 8.196\).
This value lies between 8 and 8.3.
Revision Table: Key Trigonometry Concepts
| Concept | Description | Relevant in this Problem |
|---|---|---|
| Angle of Depression | Angle between the horizontal line and the line of sight downwards. | Used to determine angles of elevation. |
| Angle of Elevation | Angle between the horizontal line and the line of sight upwards. | Equal to the angle of depression from the same horizontal level to the same point. Used in right triangles. |
| Tangent Function | In a right triangle, \(\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}\). | Used to relate the height of the tower to the distance of the car from the base. |
| Uniform Speed | Constant speed, meaning distance = speed \(\times\) time. | Used to relate the distances traveled by the car in different time intervals. |
Additional Information: Applications of Trigonometry
Trigonometry, especially the concepts of height and distance, is widely used in various fields:
- Navigation: Calculating distances and directions.
- Surveying: Measuring land boundaries and elevations.
- Engineering: Designing buildings, bridges, and other structures.
- Physics: Analyzing wave phenomena, projectile motion, and forces.
- Astronomy: Calculating distances to stars and planets.
Problems involving angles of elevation and depression are common applications that help us determine heights or distances indirectly using angles and known lengths.
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