Your vessel is steering 283°T at 10 knots....

Question 1 of 2707066636ee928f7522a1c51a65a

Your vessel is steering 283°T at 10 knots. At 0538 a light bears 350°T, and at 0552 the same light bears 002°T. What will be your distance off abeam?

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Question 1 of 27070

Your vessel is steering 283°T at 10 knots. At 0538 a light bears 350°T, and at 0552 the same light bears 002°T. What will be your distance off abeam?

🔍 Key Concepts

• Finding distance off abeam using relative bearings taken over time while on a constant course and speed • Using the run between two bearings and the change in bearing angle to form a right triangle with the distance off as one side • Relationship between time, speed, and distance: $\text{Distance} = \text{Speed} \times \text{Time}$

💭 Think About

• How much time passed between the first and second bearing, and how far did the vessel travel in that time at 10 knots? • What is the change in bearing from 350°T to 002°T, and how does that change relate to the geometry of the track line and the light’s position? • If you draw the vessel’s track as a straight line and plot the two lines of bearing from the light, what kind of triangle is formed, and which side of that triangle represents the distance off abeam?

✅ Before You Answer

• Confirm elapsed time between 0538 and 0552, then convert that to distance run at 10 knots. • Check that you’re using the change in bearing (from 350°T to 002°T), not the absolute bearings, to set up the geometry. • Verify that you correctly identify the side of the triangle that corresponds to the distance off abeam (the perpendicular distance from your track to the light).