Question

An observer 1.6 m tall is 20_3 away from a tower. The angle of elevation from his eye to the top of the tower is 30. The heights of the tower is:

Answer 1

Answer:

19 metres

Step-by-step explanation:

if the distance away from tower is 20√3

then,

height of tower = height of observer + height of tower to eyesight

sin30 = height of tower to eyesight / distance from tower

1/2=h of tower/ 20√3

height of eyesight = 10√3

now,

height of tower = height of observer + height of tower to eye sight

height of tower = 1.6 + 10√3

h of tower = 1.6 + 10 * 1.73 (√3=1.73)

h = 1.6+17.3

h = 18.9

h= 19 m