Question

A laptop 5 metre high is placed on a building 25 m high if the plan and building both substance equal angle on the observer at the height of 30 m then the distance between observer and the top of the flag is

Answer 1

Answer:

Step-by-step explanation:

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.