Question

An observer, 1.6m tall, is 45 metres away from a tower the angle of elevation from s eye to the top of the tower is 30 degre determine the height of the tower [take √3= 1.732]

Answer 1

Let AB = h,

In ΔABC
Tan 30 = AB/BE

1/√3 = h/45
h = 45/√3  x  √3/√3
h = 45√3/3
h = 15√3
h = 15 (1.732)
h = AB = 25.98
h ≈ 26m.----------(i)

Now AC = AB + CB

CB = 1.6m
AC = 26 + 1.6 (from i)
AC = 27.6m

Answer 2

Answer:

here is ur answer

Step-by-step explanation:

Let AB = h,

In ΔABC

Tan 30 = AB/BE

1/√3 = h/45

h = 45/√3  x  √3/√3

h = 45√3/3

h = 15√3

h = 15 (1.732)

h = AB = 25.98

h ≈ 26m.----------(i)

Now AC = AB + CB

CB = 1.6m

AC = 26 + 1.6 (from i)

AC = 27.6m