1. ## Sin and cos rules work , help please

hi.
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 2am the bearings of the lighship are 149degrees and 142degrees.

I have draw a diagram to help the problem, and i have tried a few ways of finding the answers but i am soo lost , can you please help i need to;
find the distance of the ship fromA at 2am, and also find the speed of the ship - the speed i dont know where to begin

thank you

2. 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 $P.$
Two hours later, it is at $Q.$
. . $\angle NQA = 149^o\quad\Rightarrow\quad \angle AQP = 31^o$
. . $\angle PQB = 142^o\quad\Rightarrow\quad \angle BQA = 7^o$
Let $x = QA,\;y = QP$

In right triangle $QPA,\;\angle AQP = 31^o\quad\Rightarrow\quad \angle QAP = 59^o \quad\Rightarrow\quad \angle QAB = 121^o$

In triangle $QAB,\;\angle B \:=\:180^o - 7^o - 121^o \quad\Rightarrow\quad \angle B\:=\:52^o$

Law of Sines: . $\frac{x}{\sin52^o} \:=\:\frac{10}{\sin7^o}\quad\Rightarrow\quad x \:=\:\frac{10\sin52^o}{\sin7^o} \:=\:64.66029369$

(a) Therefore: . $x \:\approx\:64.7\text{ km}$

In right triangle $QPA,\;\cos31^o\:=\:\frac{y}{x}\quad\Rightarrow\qua d y \:=\:64.7\cos31^o \:=\:55.45872436$

Hence: . $y \:\approx\:55.5\text{ km}$

(b) Therefore, the ship's speed was: . $\frac{55.5\text{ km}}{2\text{ hrs}} \;=\;27.75\text{ km/hr}$

3. Hi, thank for that its great but i dont understand were the 121degrees has come from?
thnkz

4. its ok lol iv worked it out

thnks