The problem of doubling the angle on the bow with set due to current to find the distance from the light house at the time of second observation.

Step 1

A vessel while moving on a course steered represented by line $\bar AB$ observes the bearing of a light house at D. After a certain duration, the vessel again observes the same light house at a bearing which is double the relative bearing at position A. The $\bar BC$ represents the vessel’s drift during the duration of steaming between the two observations. The construction is done accordingly.

Step 2

From C line parallel to AB is drawn until it cuts line AD at F.CE is drawn parallel to AD until it cuts the line AB at E. AC is joined.

$$FC = CD = AE = (AB -BE)$$

and

$$\angle EAD = \angle CFD\ = \angle CEB$$

because FC parallel to AB, because AD parallel to EC. Let $\Phi$ = the relative angle between drift direction and course steered

Step 3

In $\triangle BCE$, Applying Sine formula,

$$\frac{sin(\angle(BCE))}{BE} = \frac {sin(\angle (CBE))}{EC} = \frac {sin (\angle BEC)}{BC}$$

. We know $\angle DAB = \angle BEC = \theta$ as DA is parallel to CE. We also the the distance of BC = drift of the ship during the interval between the two observations = d (assumed). Then from above equation.

$$\frac {sin (\theta)}{d}= \frac {sin \angle BCE}{BE}$$

Now

$$\angle BCE = 180\degree - \theta -( 180 \degree - \Phi )$$

because line EB extended is $180 \degree$ and in $\triangle BCE$sum of all angles$180 \degree$

$$\therefore \angle BCE = \Phi -\theta$$

$$\therefore \frac { sin (\theta)}{d}= \frac{sin(\Phi - \theta) }{BE}$$

$$\therefore BE = \frac { d \times sin(\Phi - \theta)}{sin (\theta)}$$

$$\therefore AE = AB(known) - BE (determined)$$

AE = CD = distance off the light house at the time of second observation. This was the required solution.