Question

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

Answer 1

Answer:

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

Answer 2

Answer:

Let △ABC be a right angled triangle where ∠B=90

0

and ∠DAC=45

0

as shown in the above figure.

Let x be the distance of object from the building.

Since AD∣∣BC, therefore,

∠DAC=∠BCA=45

0

(Alternate angles)

We know that tanθ=

Adjacentside

Oppositeside

=

BC

AB

Here, θ=45

0

, BC=x m and AB=50

3

m, therefore,

tanθ=

BC

AB

⇒tan45

0

=

x

50

3

⇒1=

x

50

3

(∵tan45

0

=1)

⇒x=50

3

Hence, the distance of the object from the building is 50

3

m.