Anonymous asked in Science & MathematicsMathematics · 1 month ago

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'.

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1 Answer

  • 1 month ago
    Favourite 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)

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