A ship leaves port at 1:00 P.M. and sails in the direction N37°W at a rate of 23 mi/hr. Another ship leaves po?
A ship leaves port at 1:00 P.M. and sails in the direction N37°W at a rate of 23 mi/hr. Another ship leaves port at 1:30 P.M. and sails in the direction N53°E at a rate of 15 mi/hr.
a.Approximately how far apart are the ships at 3:00 P.M.? (Round your answer to the nearest whole number.)
(b) What is the bearing, to the nearest degree, from the first ship to the second?
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Assuming that N37°W is 37° west of north or 323° and N53°E is 53° east of north or 053°, then the difference between the two headings is 90° and their paths form a right triangle.
From 1pm to 3pm, the first ship travels at 23 mi/hr for 2 hrs or 46 miles. From 1:30pm to 3pm, the second ship travels at 15 mi/hr for 1.5 hrs or 22.5 miles. These two distances are the legs on your right triangle and the distance (d) between the two of them at 3pm is the hypotenuse.
d^2 = (46)^2 + (22.5)^2 = 2116 + 506.25 = 2622.25
d = sqrt(2622.25) = 51.21 miles
d = 51 miles ANSWER to (a)
Finding the bearing between the two at 3pm is slighlty more complicated. If the relative angle between the two ships at 3pm is A, then:
sin(A) = (opposite leg)/(adjacent leg) = (22.5)/(46) = 0.4891
A = arcsin(0.4891) = 29°
Since the first ship is heading 323°, its back angle is 143°, and the bearing to the second ship is 29° back from that or 114° ANSWER to (b)
Try graphing it out, then measure the bearing to make sure I did not make any mistakes.
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