The angle of depression of a boat from the top of a cliff 10m high is 30. How far is the boat from the foot of the cliff?
- \(\frac{5√3m}{3}\)
- 5√3m
-
10√3m
- \(\frac{10√3m}{3}\)
The correct answer is: C
Explanation
RQP = 90oPQS = 30o but
RQS = 90o = \(\frac{RS}{10}\)
RS = 10 tan 60o
RS = 10 tan 60o
RS = 10√3m