Question

from the top of the building 20√3 m high , the angle of depression of an object on the ground is observed to be 30°. find the distance of the object from the building .





(question of some of application of trigonometry)

please explain the answer







and the answer should be 60m please explain ​

Answer 1

Answer:

tan30°=height of the building/distance from foot of the building

1/√3=20√3/d

d=20×3=60m

Answer 2

$\huge\bold\purple{AɴS᭄ᴡEʀ}$

Let △ABC be a right angled triangle where ∠B=90° and ∠DAC=45° as shown in the above figure. 

Let x be the distance of object from the building.

Since AD || BC, therefore

∠DAC=∠BCA=45° (Alternate angles)

We know that tanθ=Adjacentside/ Oppositeside =BC/AB

Here, θ=45°, BC=x m and AB=503 m, therefore,

$tan∅=AB/BC$

$⇒tan 45°=(50√3)/x$

$⇒1=(50√3)/x$

$(∵tan450=1)⇒x=503$

$\bold\purple{ \:Hence, the \: distance \: of \: the object\: }$

$\purple{from the building is 50√3 m.}$