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P and Q are two points observed from the top of a building 10√3 m high. If the angles of depression of the points are complementary and PQ = 20 m, then the distance of P from the building is (SSC CGL 1st Sit. 2012)
- a)25 m
- b)45 m
- c)30 m
- d)40 m
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
P and Q are two points observed from the top of a building 10√3 ...
In this type of question, just put
h2 = x(x + 20)
⇒ (10√3)2 = x (x + 20)
300 = x (x + 20)
10 (30) = x (x + 20)
x + 20 = 30
Most Upvoted Answer
P and Q are two points observed from the top of a building 10√3 ...
Given information:
- Height of the building = 10√3 m
- Distance between points P and Q (PQ) = 20 m
- Angles of depression of points P and Q are complementary
Solution:
- Let the distance of point P from the building be x meters.
Calculating distances:
- In triangle APB, tanθ = (10√3) / x
- In triangle AQB, tan(90-θ) = (10√3) / (20-x)
Using complementary angles property:
- tan(90-θ) = cotθ
- Therefore, cotθ = (10√3) / (20-x)
Substitute tanθ = (10√3) / x:
- (1 / tanθ) = cotθ
- 1 / ((10√3) / x) = (10√3) / (20-x)
- x / (10√3) = (10√3) / (20-x)
Solving for x:
- x(20-x) = (10√3)^2
- 20x - x^2 = 300
- x^2 - 20x + 300 = 0
- (x-10)(x-30) = 0
- x = 10 or x = 30
Conclusion:
- Since x cannot be 10 (as it would be inside the building), the distance of point P from the building is 30 meters. Therefore, the correct answer is option C.
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