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Old as Dirt!
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You can obtain the solution graphically by plotting the initial position and Bearing to the target and them plotting an EP taking into account the set and drift over the subsequent 30 minutes; and, the yachts heading and speed, which would give you a CMG of 199.59ºT and distance traveled of 3.21 miles and from that point lay down a line of 347º and measure the distance to the Intersect from the EP; or, by using the law of sines simply calculate the distance from the EP to the intercept which is 4.32 miles
 

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Old as Dirt!
Joined
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3,488 Posts
You can obtain the solution graphically by plotting the initial position and Bearing to the target and them plotting an EP taking into account the set and drift over the subsequent 30 minutes; and, the yachts heading and speed, which would give you a CMG of 199.59ºT and distance traveled of 3.21 miles and from that point lay down a line of 347º and measure the distance to the Intersect from the EP; or, by using the law of sines simply calculate the distance from the EP to the intercept which is 4.32 miles
The theory described above is correct although the numbers given are in error which serves to illustrate why someone should not try doing trigonometry when one is under the influence of pain-killers (Percoscet) for a recent surgery.

Recalibrating, from the initial observation point (OP1), the bearing to the mark is given as 340º. From the second observation point (OP2), the bearing is given as 347º such that the included angle between the two bearings is 7º. One must then determine the position of OP2 relative to OP1 taking into account set, drift, course and speed.

The set is given as 135º (or 45º below the horizontal) at a drift of 1.2 knots. Over the 30 minute interval between observations, the EP would move a distance of .6 miles--east .4243 miles and south .4243 miles (in each case .6 x sin 45º).

The adjusted EP to OP2 would than be taken into account based upon the yacht's heading of 240º at 6 knots or a distance of 3 miles. From the adjusted EP the yacht would move south 1.5 miles (3 x sin 30º) and west 2.598 miles (3 x cos 30º). Relative to OP1, OP2 would be positioned 1.9243 miles south (being 1.5 miles plus .4243 miles) and west 2.1738 miles (being 2.598-.4243 miles east). By the square root of the sum of the squares, the distance traveled would have been 2.9031 miles on a heading of 228.49º (270º - arcsin (1.9243/2.1738)). From OP2, the bearing to the mark is given as 347º (see above).

By the law of sines, the lengths of the sides of a triangle are proportional to the sins of the opposing angles. The objective is to find the distance from OP2 to the mark. The included angle facing the unknown side is 111.5153º (being 340º - 228.49º). The included angle between OP1 and OP2 at the mark is 7º (being 347º-340º) and the opposing side is 2.9031 miles long. And, 2.9031/sin 7º = L/sin 111.5153º. Solving for L one arrives at 22.1619 miles (being 2.9031 x sin 111.5153º/sin 7º). Rather a different answer than my original hen-scratch, eh? (Intuitively, one can tell the mark is rather a long way off given the minor change in the bearing-7º-over 30 minutes running at 6 knots.)

FWIW...
 
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