From the top of an upright pole 24√3 feet high, the angle of elevation of the top of an upright tower was 60°. If the foot of the pole was 60 feet away from the foot of the tower, what tall (in feet) was the tower?
Show answer & explanation
Correct answer: A
Explanation
The correct answer is Option 1 i.e. 84√3
Let the height of a tower CD be h feet
The distance from a pole AB to a tower CD = 60feet
tan 60º = (h - 24√3)/60
√3 = (h - 24√3)/60
h = 84√3 feet