The angles of depression of the top and the bottom of a building 50

metres high as observed from the top of a tower are and  respectively. Find the height of the tower and also the horizontal distance between the building and the tower.

Asked by Topperlearning User | 3rd Nov, 2017, 12:52: PM

Expert Answer:

 

      begin mathsize 12px style Let space AB thin space be space the space building equals 50
CE space be space the space tower equals 50 plus straight y
Let space BC equals AD equals straight x equals the space distance space between space them
tan 30 to the power of 0 equals DE over AD
rightwards double arrow fraction numerator 1 over denominator square root of 3 end fraction equals straight y over straight x end style

      begin mathsize 12px style In space right space increment EBC comma
tan 60 degree space equals space CE over BC
square root of 3 equals fraction numerator 50 space plus space straight y over denominator straight x end fraction
square root of 3 straight x space equals space 50 space plus space straight y
square root of 3 open parentheses square root of 3 straight y close parentheses space equals space 50 space plus space straight y........ from space left parenthesis straight i right parenthesis
3 straight y space minus space straight y space equals space 50
2 straight y space equals space 50
straight y space equals space 25
Height space of space the space tower space CE space equals space 50 space plus space straight y
equals 50 space plus space 25
equals space 75 space cm
Now space taking space the space value space of space straight y space in space left parenthesis straight i right parenthesis comma space we space get
straight x space equals space square root of 3 cross times 25
straight x space equals space 1.73 cross times 25
straight x space equals space 43.25
Required space horizontal space distance space straight x space equals space 43.25 space straight m
end style

          

Answered by  | 3rd Nov, 2017, 02:52: PM