Trigonome-FUN... Come one, come all.

I've got a question that is probably not to difficult accomplish, I just can't think out of the box anymore. I'll list below this first question what I've done so far.

(1)

This is like the preceding problem, except that the numbers are a little different. A ship is moving due west at 5 knots. You are in a speed boat at a distance of 7 nautical miles from the ship. Your bearing as seen from the ship is 102 degrees. You need to catch up with the ship. So you take off at a speed of 17 knots and a bearing of __________(a) degrees, and you reach the ship in ___________(b) minutes. Enter your answers as decimal expression with at least four digits, or enter mathematical expressions.

ANS:

(a) 278.7676399

t=time or rather the knotts variable

5t+7cos(12)

..\ ----------

...\.............|

....\............|

17t.\..........| 7sin(12)

.......\........|

.........\......|

...........\....|

............\...|

(*Yeah it's a crappy ASCII triangle, but I was going for understanding here...)

thus

(5t+7cos(12))^2+(7sin(12))^2=17t^2

or

264t^2-70tcos12-49=0

Through that into the quadratic

-bħsqrt(b^2-4(a)(c))/ 2(a) and you'll get t. Of which, I got the following answer.

t = 0.1524276704

Which is in hours, and I need it in minutes... So I calculate t(60) and get the following. 9.145660224 minutes, which doesn't work when I enter it in. Any help would be greatly appreciated. :D