Trigonometric bearing problem I have two trigonometric problems that I solved, however it does not match the answer in the book:


  1) A yacht crosses the start line of a race on a bearing of $31$ degrees. After $4.3$ km, it rounds a buoy and sails on a bearing of $346$ degrees. When it is due north of its start, how far has it sailed altogether. 


I calculated the answer to be $6.353$ km. However the book's answer is $13$ m. The units are themselves different.


  2) The bearing of $B$ and $A$ is $65$ degrees. The bearing of $C$ from $B$ is $150$ degrees, and the bearing of $A$ from $C$ is $305$ degrees. If $AC=300$m, find $BC$.


I calculated the answer to be $-1333.838$. However the answer in the book was $261$ km.  The units are themselves different as well.


  3) The bearing of $Y$ from $X$ is $205$ degrees. The bearing of $Z$ from $Y$ is $315$ degrees, and the bearing of $X$ from $Z$ is $85$ degrees. If $XY = 4$ km, find the distance $XY$.


I calculated the answer to be $3.26$ km.

The questions are from Edexcel IGCSE Ex-181* 
Please check if my answers are correct? If not please show the steps to get the correct answer. I used the sine rule to solve these sums.
The boat leaves at a bearing of 31 degrees and ends up north of it's start point. Therefore first corner of the triangle is 31 degrees.

At the buoy it has a bearing of 346 which is NW - ish. This angle is equal to 135 degrees.
1st part of angle is 90-31= 59degrees using construction lines and the alternate angles theorem. The second part of the angle is 346-270=76 degrees. (the bearing minus the East-West construction line) so 76+59=135 degrees.

The last angle is 180-135-31=14 degrees.

I now use Sine rule: x/sin135  =  4.3/sin 14. x = 12.57 - which rounds to 13km.
This is the answer for the question no2

this is the answer for the question no3