Respuesta :
30.8 hrs is the same 15-knot speed is maintained.
What is the distance?
- The extent or amount of space between two things, points, lines, etc. the state or fact of being apart in space, as of one thing from another; remoteness. a linear extent of space.
- Seven miles is a distance too great to walk in an hour. an expanse; area: A vast distance of water surrounded the ship.
Distance from SJ to location of ship can be determined since we have rate and time
D= rate x time = 15 knots x 10 hrs = 150 nautical miles
Since we have no other angles, we cant use law of sines yet. We have a SAS now, so apply law of cosine
S to B = [tex]\sqrt{600^{2} + 150^{2} + - 2(600)(150)cos(20) }[/tex]
Ship to Barbados = 461.9 nauticalmiles
Now we can use law of sines to determine the angle.
[tex]\frac{461.9}{sin20} = \frac{600}{sin 0}[/tex]
sinθ = [tex]\frac{600sin20}{461.9}[/tex]
θ = 26.4 degrees
But this is the angle within the triangle, so 180 - 26.4 =153.6 degrees
b) We already determined the distance to Barbados in part awhen we used law of cosine. Distance is 461.9 nautical miles
Rate is 15 knots
D = R*T
T= D/R
T = 461.9/15
T = 30.8 hrs
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The complete question is -
A cruise ship maintains an average speed of 15 knots in going from San Juan, Puerto Rico, to Barbados, West Indies, a distance of600 nautical miles. To avoid a tropical storm, the captainheads 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 the storm.(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 same 15-knot speed is maintained? See picture below. Let B = Barbados Let SJ = San Juan Let S = location of ship
