Trigonometry question 2?
There is a vertical cliff (75m high). At a particular instant 't', the angle of depression of a ship viewed from the top of the cliff is 30◦. The ship is sailing away from the cliff at constant speed and 1 minute later its angle of depression from the top of the cliff is 20◦. Determine the speed of the ship in 'm/s' and in 'km/h'.
- llafferLv 71 month agoFavourite answer
distance = rate * time
The distance here is the change is positions between the two checks (currently unknown)
rate is unknown (what you are asked to solve for)
time from point A to point B is 1 minute.
We have two right triangles with the same 75 m height, unknown bases, and different angles. We can use tangents to solve for the two bases then subtract them to get the distance traveled in the minute:
tan() = opp/adj
tan(30) = 75 / x₁ and tan(20) = 75 / x₂
Solve for both unknowns:
x₁ = 75 / tan(30) and x₂ = 75 / tan(20)
x₁ = 129.9038 and x₂ = 206.0608 (rounded to 4DP)
The difference is:
206.0608 - 129.9038 = 76.157 m
That's the distance traveled in 1 min so the rate is:
d = rt
76.157 = r(1)
r = 76.157 m/min
You are asked to give the rate in meters per second and km per hour so we convert:
76.157 m/min (1/60 min/sec) = 1.269 m/sec (rounded to 3DP)
76.157 m/min (60 min/hr) (1/1000 km/m) = 4.569 km/h (rounded to 3DP)