兩條極難o既vector問題~(20PT)

(1) Since the 3 forces are in equilibrium, that means resultant is zero, so the 3 vectors will form a triangle with their magnitude3 as the 3 sides. So using the Cosine Rule, 1220^2 = 980^2 + 760^ - 2(980)(760) cos x
x = arc cos (0.03329) = 88.1 degree.
similarly, 980^2 = 1220^2 + 760^2 - 2(760)(1220) cos y
y = arc cos (0.5962) = 53.4 degree.
Remaining angle = 180 - 88.1 - 53.4 = 38.5 degree.
[Note: You can also use the Sine Rule in finding the 2nd angle).
(2) Question not clear.

2009-12-15 22:02:17 補充：
Let A be the angle the plane makes with the x - axis, so position vector of the plane is
168 cos A i + 168 sin A j.
Position vector of the wind = 27.1j.

2009-12-15 22:07:13 補充：
Let ground speed be s and 57 degree 40 sec is the true bearing, so resultant vector is s cos(90 - 57 deg 40 sec) i + s sin ( 90 - 57 deg 40 sec)j. That means
168 cos A = s cos ( 90 - 57 deg 40 sec) = 0.845s ..... (1) and 168 sin A + 27.1 = s sin ( 90 - 57 deg 40 sec) = 0.5348s .... (2)

2009-12-15 22:12:01 補充：
Eliminating s from (1) and (2) and doing some transformation we get 371.765 cos (A + 57.67) = 50.673. So A = arc cos(50.673/371.765) - 57.67 = 24.5 degree ( True bearing is 065.5). Sub into (1) we get s = 180.1 mph.

2009-12-15 22:50:59 補充：
Alternatively you can eliminate A first by using sin^2 A + cos^2 A = 1 to form a quadratic equation in s.