4200.23 – The Yacht Gemini on the Hudson

Tags:Problem Set 12Pre-Calculus MathTrigonometry

You are sailing your splendid yacht Gemini, formerly owned by Edward, Duke of Windsor, down the Hudson River into New York Harbor. It is a sunny afternoon, with a light breeze, and everything is going well. You are approaching the George Washington Bridge. Beneath the bridge, on the Manhattan side, to your left, is a small red lighthouse. You take a sighting and observe that this lighthouse is 15 dgrees to port. You continue downriver for two minutes at your stately rate of 5 knots, and you observe that the lighthouse isnow 29 degrees to port. How close will you come to the lighthouse as you pass under the bridge?

(N.b. 1 knot = 6076 feet or 6080.27 feet--sources vary.)


A is the position of Gemini when you spot the lighthouse (L).

B is the boat's position 2 minutes later.

The distance xx is the closest you'll come to the lighthouse.

aa is the distance the boat travels in 2 minutes.

kk is the distance BL.

If we can find kk, we can use BCL\triangle BCL to find xx.

If we can find ALB\angle ALB, we can find kk by using the law of sines.

That's not so hard to do: \begin{aligned} \angle ABL = 180n- 29 - 151.\ \angle ALB = 180 - 151 - 15 = 29 - 15.\ \angle ALB = 14° \end{aligned}

So: we still need the distance aa: a=2605a = \frac{2}{60}\cdot5 knots = 16\frac{1}{6} knot = 607661013\frac{6076}{6} \approx 1013 feet.

Now then: asin14=ksin15k=1013sin15sin141084\frac{a}{sin 14} = \frac{k}{sin 15} \rightarrow k = \frac{1013sin 15}{sin 14} \approx 1084 feet.

So x1084=sin29x=1084sin29525feet\frac{x}{1084} = sin 29 \rightarrow x = 1084sin 29 \approx 525 feet.

We must note that the sketch is inaccurate in that it shows Gemini more than halfway to the New Jersey side of the river.

GL, the width of the river, is actually more than 4000 feet. 525 feet? Don't run aground!

If you do a Google image for Elco yacht Gemini you can see a picture of it (its original name was Serenity; it was re-named for the Dutchess's astrological sign).