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An airplane flies 200 km due west from city A to city B and then 280 km in the direction of 30.5° north of west from city B to...Asked by Jane
An airplane flies 200 km due west from city A to city B and then 240 km in the direction of 31.0° north of west from city B to city C.
(a) In straight-line distance, how far is city C from city A?
(b) Relative to city A, in what direction is city C?
(a) In straight-line distance, how far is city C from city A?
(b) Relative to city A, in what direction is city C?
Answers
Answered by
Damon
Law of cosines
b^2 = a^2+c^2-2ac cos B
b^2 = 240^2 + 200^2 -2(240)(200)cos(180-31)
solve for b which is distance from A to C
then law of sines
sin A/240 = sin(180-31)/b
solve for angle A
b^2 = a^2+c^2-2ac cos B
b^2 = 240^2 + 200^2 -2(240)(200)cos(180-31)
solve for b which is distance from A to C
then law of sines
sin A/240 = sin(180-31)/b
solve for angle A
Answered by
Anonymous
As a sailboat sails 59.0 m due north, a breeze exerts a constant force 1 on the boat's sails. This force is directed at an angle west of due north. A force 2 of the same magnitude directed due north would do the same amount of work on the sailboat over a distance of just 36.0 m. What is the angle between the direction of the force 1 and due north?
Answered by
Cory
because it says the two forces do the same amount of work, you can tell that the two distances given are the hypotenuse and leg of a right triangle. after that it's just a matter of taking the inverse cos of adjacent/hypotenuse
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