Question

solve this Q of lesson heights and distances
pls plz solve this
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Answer 1

Step-by-step explanation:

Let the height of the man = AD = 2 m.

∴ CE = 10 - 2 = 8 m.

(i)

In ΔCED,

CE/DE = tan x° = 2/5

⇒ (8/DE) = 2/5

⇒ 40 = 2 * DE

Here, AB = DE

∴ AB = 20 cm

(ii)

Let A'D be the new position of man and θ be the angle of elevation of the top of tower.

So,

D'E = 15 m

In ΔCED,

tan θ = CE/D'E

tan θ = 8/15

tan θ = 0.533

θ = 28°

Therefore,

Angle of elevation θ = 28°

Hope it helps!