Trigonometry question 2?

Question

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'.

Answer 1

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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)