Tejas is standing at the top of a building and observes a car at an angle of depression of 30° as it approaches the base of the building at a uniform speed. 6 seconds later, the angle of depression increases to 60", and at that moment, the car is 25 m away from the building.
Question: 1
What is the height of the building ?
Show Hint
Draw a clear diagram. The angle closer to the base of the building is always the larger angle (60\(^\circ\) in this case).
Step 1: Understanding the Concept:
Angle of depression is equal to the angle of elevation due to alternate interior angles. We use trigonometric ratios in the right-angled triangle. Step 2: Key Formula or Approach:
In \( \triangle ABC \) (where A is the top and BC is the base), \( \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} \). Step 3: Detailed Explanation:
Let the height of the building be \( h \).
At \( 60^\circ \) depression, the car is at point C, 25 m from the base B.
In right \( \triangle ABC \):
\[ \tan 60^\circ = \frac{AB}{BC} \]
\[ \sqrt{3} = \frac{h}{25} \]
\[ h = 25\sqrt{3} \text{ m} \] Step 4: Final Answer:
The height of the building is \( 25\sqrt{3} \) m.
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Question: 2
What is the distance between the two positions of the car ?
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Distance CD is found by calculating the full horizontal distance from the base and subtracting the known partial distance.
Step 1: Understanding the Concept:
We need to find the distance CD between the car's initial position (D) at \( 30^\circ \) and its second position (C) at \( 60^\circ \). Step 2: Detailed Explanation:
In right \( \triangle ABD \):
\[ \tan 30^\circ = \frac{AB}{BD} \]
\[ \frac{1}{\sqrt{3}} = \frac{25\sqrt{3}}{BD} \]
\[ BD = 25\sqrt{3} \times \sqrt{3} = 25 \times 3 = 75 \text{ m} \]
Distance between positions (\( CD \)) = \( BD - BC \)
\[ CD = 75 - 25 = 50 \text{ m} \] Step 3: Final Answer:
The distance between the two positions of the car is 50 m.
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Question: 3
What would be the total time taken by the car to reach the foot of the building from the starting point ?
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Alternatively, since distance 50m takes 6s, 25m takes 3s. Total time = 6 + 3 = 9 seconds.
Step 1: Understanding the Concept:
First, find the uniform speed of the car using the distance it covered in 6 seconds. Then find the total time for the total distance. Step 2: Detailed Explanation:
Distance \( CD = 50 \) m was covered in 6 seconds.
Speed \( v = \frac{\text{Distance}}{\text{Time}} = \frac{50}{6} = \frac{25}{3} \text{ m/s} \).
The total distance from starting point D to foot B is \( BD = 75 \) m.
Total time taken = \( \frac{\text{Total distance } BD}{\text{Speed}} \)
\[ \text{Time} = \frac{75}{25/3} = \frac{75 \times 3}{25} = 3 \times 3 = 9 \text{ seconds} \] Step 3: Final Answer:
The total time taken is 9 seconds.
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Question: 4
What is the distance of the observer from the car when it makes an angle of \( 60^\circ \) ?
Show Hint
You can also use Pythagoras theorem \( h^2 + 25^2 = AC^2 \) but using cosine is much faster.
Step 1: Understanding the Concept:
The distance of the observer (at A) from the car (at C) is the length of the hypotenuse AC in \( \triangle ABC \). Step 2: Key Formula or Approach:
Use \( \cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} \). Step 3: Detailed Explanation:
In right \( \triangle ABC \):
\[ \cos 60^\circ = \frac{BC}{AC} \]
\[ \frac{1}{2} = \frac{25}{AC} \]
\[ AC = 25 \times 2 = 50 \text{ m} \] Step 4: Final Answer:
The distance of the observer from the car is 50 m.
