JAMB - Mathematics (1994 - No. 30)
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
\(\frac{10√3m}{3}\)
Explication
RQP = 90o
PQS = 30o but
RQS = 90o = \(\frac{RS}{10}\)
RS = 10 tan 60o
RS = 10 tan 60o
RS = 10√3m
PQS = 30o but
RQS = 90o = \(\frac{RS}{10}\)
RS = 10 tan 60o
RS = 10 tan 60o
RS = 10√3m