What is distance of ship from coastline(AB) if it is 167° and 205° from A and B respectively?

Question

Two points A and B on a straight coastline are 1km apart, B being due east of A. If a ship is observed on bearings 167° and 205° from A and B respectively, what is its distance from the coastline?

Any tips?

I ended up with at triangle containing the degrees 142°, 13° and 25° and then using the sin rule to find sides BC and AC. But it already feels like I'm going in the wrong direction since I'd have to get the perpendicular of C which would make two triangles and then use pythagoras AND simultaenous equations. Seems wrong.

Answer 1

Your triangle approach certainly could work (save for any careless mistake).

Personally, I would just define $A=(0,0)$ and $B=(1,0)$ and define two lines \begin{align} x&=y\tan(167^\circ)\\ x-1&=y\tan(205^\circ) \end{align}

and find the intersection via \begin{align} y\tan(205^\circ)+1&=y\tan(167^\circ)\\ y(\tan(205^\circ)-\tan(167^\circ))&=-1\\ y&=\frac{-1}{\tan(205^\circ)-\tan(167^\circ)} \end{align}

You can see the graph to help you see better.

Answer 2

I would use the Law of Sines to find another side. Then you can find the area of the triangle from two sides and the included angle (i.e. $\frac{1}{2}a b \sin C$). Finally, you can solve for the distance to shoreline because it is the altitude to the base of 1km.