Posted by Dakota S on Sunday, September 12, 2010 at 9:08pm.
(a) Add the vectors of his 55 mile and 25 miles flights. The reverse of that vector is the direction and distance he must fly the next day.
(b) Add the magnitudes of the three flight vectors in the previous question.
15 deg. E of S = 285 deg CCW from 0 deg
X = HOR = 25cos(285) - 55 = -48.5,
Y = VER = 25sin(285) = -24.15,
tanA = Y/X = -24.15/-48.5 = 0.4979,
A = 26.47 deg. = 206.47 deg.(3rd quad.)
d = Y/sinA = -24.15/sin(206.5)=54.18mi
d = 54.18 mi @ 206.47 deg.
a. 54.18 mi N @ 63.5 deg W of South
Location: 3rd QUADRANT.
B. D = 2 * 54.18 = 108.36 MILES
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