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September 30, 2014

September 30, 2014

Posted by **Dakota S** on Sunday, September 12, 2010 at 9:08pm.

- Physics -
**drwls**, Monday, September 13, 2010 at 1:38am(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.

- Physics -
**Henry**, Monday, September 13, 2010 at 2:41am15 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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