Question

if the angle of depression of an object from a 75m high tower is 30o,then the distance of the object from the base of tower is​

Answer 1

The distance of the object from the base of the tower is 129.9 m.

Step-by-step explanation:

Since we have given that

Angle of depression = 30°

Length of tower = 75 m

We need to find the distance of the object from the base of tower:

So, we will "Tangent of triangle":

$\tan 30^\circ=\dfrac{75}{Base}\\\\\dfrac{1}{\sqrt{3}}=\dfrac{75}{Base}\\\\Base=75\sqrt{3}=129.90\ m$

Hence, the distance of the object from the base of the tower is 129.9 m.

# learn more:

The angle of depression of an object from the top of a tower of height 75m is 30degree.Then the distance of the object from the foot of the tower is:

https://brainly.in/question/6585416

Answer 2

Step-by-step explanation:

Considering the ΔAOB

Angle of depression = ∠AOB (Alternate angles are equal)

Since, ΔAOB is a right angled triangle with ∠B=90°,

tan30° = AB

OB

∴ 1 = 75

√3 OB

∴ OB = 75×√3

i.e. OB = 75√3 m

Hence, the distance of the object from the tower is 75√3 m.