Question

2.*
1 point
एक वस्तु का एक 75m ऊंचे टाव
र से अवनमन कोण 30° है, तो उसकी
टावर से दूरी होगी?
If angle of depression of an object from a 75 m high tower is 30°,
then the distance of the object from the tower is?
A. 25/3m
B. 50/3m
C.75/3m
D. 150m
OA​

Answer 1

Given : angle of depression of an object from a 75 m high tower is 30°,

To Find :  the distance of the object from the tower

Solution:

angle of depression = angel of elevation

Tan  angle of elevation =  Height of tower /  horizontal  distance of the object from the tower

=> Tan 30° =  75 / horizontal  distance of the object from the tower

=> 1/√3 =  75 / horizontal  distance of the object from the tower

=> horizontal  distance of the object from the tower  = 75√3  m

horizontal  distance of the object from the tower   = 75√3  m

Sin angle of elevation =  Height of tower /  distance of the object from the top of the tower

=>  Sin 30° =  75 / distance of the object from the top of the tower

=> 1/2 =  75 / distance of the object from the top of the tower

=>distance of the object from the top of the tower = 150  m

Learn More:

From a point on the ground, the angles of elevation of the bottom ...

https://brainly.in/question/15934671

The angles of elevation of the top of a lighthouse from 3 boats A , B ...

https://brainly.in/question/17777927