Question

from the top of a vertical building of 50√3m height on a level ground the angle of depression of an object on the same ground is observed is to be 60°.find the distance of the object from the foot of the building​

Answer 1

Step-by-step explanation:

 △ABC be a right angled triangle where ∠B=900 and ∠DAC=450 as shown in the above figure. 

Let x be the distance of object from the building.

Since AD∣∣BC, therefore,

∠DAC=∠BCA=450    (Alternate angles)

We know that tanθ=AdjacentsideOppositeside=BCAB

Here, θ=450, BC=x m and AB=503 m, therefore,

tanθ=BCAB⇒tan450=x503⇒1=x503(∵tan450=1)⇒x=503    

   

Hence, the distance of the object from the building is 503 m.