a flagstaff 5 m high is placed on a building 25 m high if the flag and building both subtend equal angles on the observer at the height 30m the distance between observer and the top of the flag is
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Answer:
5√(3/2) m or 6.124 m.
Step-by-step explanation:
Image is attached, relate the steps with the image.
Given, ∠DAE = ∠ DAC = Θ (Let's assume)
Thus, ∠EAC = 2Θ
Let's assume the distance between the building and the observer is 'x' m.
Thus, for the triangle ΔACE;
EC = ED + DC = 5+25 = 30m
Opposite/adjacent = EC/EA = 30/x = tan2Θ
For the ΔEDA;
Opposite/adjacent = ED/EA = 5/x = tanΘ
Now, tan2Θ = 2tanΘ / (1-tan²Θ)
or, 30/x = 2(5/x) / (1 - 25/x²) [As tanΘ = 5/x and tan2Θ = 30/x]
or, 3 = 1 / (1-25/x²)
or, 3x² - 75 = x²
or, 2x² = 75
or, x = 5√(3/2) = 6.124
Thus the distance between the building and the observer is 5√(3/2) m or 6.124 m.
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