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
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?
3 answers
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?
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
1 answer