Hello, Chez!
At midnight a ship sailing due north passes two lightships, A and B,
which are 10km apart in a line due east from the ship.
Lighship A is closer to the ship than B.
At 2 am the bearings of the lighship are 149° and 142°.
(a) Find the distance of the ship from A at 2am.
(b) Find the speed of the ship. Code:
N


Q *
 * *
31°* 7°*
 * *
y  * x *
 * *
 * 52° *
P *       *       *
A 10 B
At midnight, the ship is at $\displaystyle P.$
Two hours later, it is at $\displaystyle Q.$
. . $\displaystyle \angle NQA = 149^o\quad\Rightarrow\quad \angle AQP = 31^o$
. . $\displaystyle \angle PQB = 142^o\quad\Rightarrow\quad \angle BQA = 7^o$
Let $\displaystyle x = QA,\;y = QP$
In right triangle $\displaystyle QPA,\;\angle AQP = 31^o\quad\Rightarrow\quad \angle QAP = 59^o \quad\Rightarrow\quad \angle QAB = 121^o$
In triangle $\displaystyle QAB,\;\angle B \:=\:180^o  7^o  121^o \quad\Rightarrow\quad \angle B\:=\:52^o$
Law of Sines: .$\displaystyle \frac{x}{\sin52^o} \:=\:\frac{10}{\sin7^o}\quad\Rightarrow\quad x \:=\:\frac{10\sin52^o}{\sin7^o} \:=\:64.66029369$
(a) Therefore: .$\displaystyle x \:\approx\:64.7\text{ km}$
In right triangle $\displaystyle QPA,\;\cos31^o\:=\:\frac{y}{x}\quad\Rightarrow\qua d y \:=\:64.7\cos31^o \:=\:55.45872436$
Hence: .$\displaystyle y \:\approx\:55.5\text{ km}$
(b) Therefore, the ship's speed was: .$\displaystyle \frac{55.5\text{ km}}{2\text{ hrs}} \;=\;27.75\text{ km/hr}$