From the top of an upright pole 17.75 m high, the angle of elevation of the top of an upright tower was 60⁰. If the tower was 57.75 m tall, how far away (in m) from the foot of the pole was the foot of the tower?
Show answer & explanation
Correct answer: D
Explanation
The correct answer is Option 4 i.e. 40√3/3.
Let the foot of the pole AB be x meters from the foot of the tower CD
tan 60º = (57.75 - 17.75)/x
⇒ x = 40/√3 = 40√3/3 m