Your vessel is on course 093°T at 15 knots....

Question 1 of 2707066636ee928f7522a1c51a5a8

Your vessel is on course 093°T at 15 knots. At 1835 a light bears 136°T, and at 1857 the same light bears 170°T. What was your distance off the light at 1857?

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Question 1 of 27070

Your vessel is on course 093°T at 15 knots. At 1835 a light bears 136°T, and at 1857 the same light bears 170°T. What was your distance off the light at 1857?

🔍 Key Concepts

• Relative motion on a straight track when taking two bearings of the same object from a moving vessel • Using distance = speed × time to find how far you travel between the bearings • The geometry of a vessel’s track and two bearings to find distance off at abeam or later bearing

💭 Think About

• First, convert the time between the two bearings into hours and find how many miles you traveled between 1835 and 1857. What is that distance? • Sketch your course line (093°T) and plot the two bearings (136°T and 170°T) from your track. How does the change in bearing help you relate your track distance to your closest distance to the light? • Which trigonometric function (sine, cosine, or tangent) naturally relates the distance along your track to the perpendicular distance off, given the angle between your course and the bearing at the later time?

✅ Before You Answer

• Be sure you use true bearings plotted correctly from your track line, not from the light • Confirm you converted 22 minutes into hours correctly before using speed–distance–time • Double-check which angle you are using in your trig function; it should be the angle between your course (093°T) and the bearing at 1857 (170°T), not the full bearing itself