a TV tower stands vertically on a bank of a Canal from a point on the other bank directly opposite the tower the angle of elevation of the top of the tower is 60° from. 20 m away from this point on the same bank the angle of elevation of the top of the tower is 30° find the height of the tower and the width of the canal
Answers
Answer:
Width of canal = 10 m
Height of tower = 17.321 m
Step-by-step explanation:
Let the width of the canal be x.
The total distance from the point 20m away from the opposite bank to the other bank will be : x + 20
We have two right angled triangles with the following properties :
1) Base = x + 20, adjacent angle = 30, height = h
2) Base = x, adjacent angle = 60, height = h
They have a common height.
We will use Tangent to form two equations as follows :
1) Tan 30 = h/(x + 20)
h = (x + 20) Tan 30.............i)
2) Tan 60 = h/x
h = x Tan 60.............ii)
Since in both cases h is equal we equate i and ii.
(x + 20) Tan 30 = x Tan 60
(x + 20) / x = Tan 60 / Tan 30
Tan 60 = 3^½
Tan 30 = (3^½)/3
Substituting we have :
(x + 20) / x = 3^½ × 3/3^½
(x + 20)/x = 3
x + 20 = 3x
3x - x = 20
2x = 20
x = 10
Lets get the value of h now:
h = 10 × 3^½ = 17.321 m
The width of the canal = 10 m
The height of the tower = 17.321 m