A cruise ship maintains an average speed of 15 knots in going from San Juan, Puerto Rico, to Barbados, West Indies, a distance of 600 nautical miles. To avoid tropical storm, the captain heads out of San Juan in a direction of 20 degrees off a direct heading to Barbados. The captain maintains the 15 knot speed for 10 hours, after which time the path to Barbados becomes clear of storms. (a) Through what angle should the captain turn to head directly to Barbados? (b) Once the turn is made how long will it be before the ship reaches Barbados if the 15 knot speed is maintained?

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A cruise ship maintains an average speed of 15 knots in going from San Juan, Puerto Rico, to Barbados, West Indies, a distance of 600 nautical miles. To avoid tropical storm, the captain heads out of San Juan in a direction of 20 degrees off a direct heading to Barbados. The captain maintains the 15 knot speed for 10 hours, after which time the path to Barbados becomes clear of storms. (a) Through what angle should the captain turn to head directly to Barbados? (b) Once the turn is made how long will it be before the ship reaches Barbados if the 15 knot speed is maintained?

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|dw:1432783836686:dw| S = san juan B = barbados M = middle point of the journey
if the ship is going 15 knots for 10 hours, then how far does the ship travel (assuming it travels in a straight line)?

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hint: 1 knot = 1 nautical mile per hour
600*15=9000 nautical miles?
is that right? or am I misinterpreting?
distance = unknown speed = 15 knots time = 10 hours distance = speed * time
OHHH ok so then it would be 150
yeah so the distance from S to M is 150 nautical miles |dw:1432784489817:dw|
what we need is the distance from M to B. We can find that by using the law of cosines
|dw:1432784585295:dw|
c^2 = a^2 + b^2 - 2ab*cos(C) c^2 = 600^2 +150^2 - 2*600*150*cos(20) ... ... ... c = ??
okay gimme a sec to calculate
I got about 555.92
that's too large
you need to be in degree mode
I changed it to degree mode and it's still pretty big, I got about 461.90
better |dw:1432785182237:dw|
then you'll use the law of sines to find angle SMB |dw:1432785226747:dw|
so sinA/a=sinB/b=SinC/c right?
yes you'll use that
S=20 degrees B= M= s=461.90402 b=150 m=600
@jim_thompson5910 are those the right values?

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