Asked by Yes
7 m/s on a bearing of 30° and 2 m/s from a bearing of 343°. How would you find the direction and magnitude using both Sine and Cosine Law? I know how to answer this question using Components....
Answers
Answered by
Damon
NOT a BEARING, they mean a HEADING
(mathematicians drive me nuts).
A bearing is the direction you look, not the drirection you head.
Anyway:
Of couse do it with components but
Call angle at origin A, 0 deg is north
it is 30 + (360-343) = 30 + 17 = 47 deg
call Angle in NW quadrant B (quad3) and Angle in NE quadrant(quad1) = C
now
BC^2 = AB^2 + AC^2 -2 AB AC cos 47
but
AB = 2 and AC = 7
so
BC^2 = 4 + 49 - 2 (2)(7)cos 47
solve for BC, now have all 3 sides
then use law of sines to get angles
like
sin C /2 = sin A/BC and Sin B / 7 = same thing
then you have all the sides and all the angles.
(mathematicians drive me nuts).
A bearing is the direction you look, not the drirection you head.
Anyway:
Of couse do it with components but
Call angle at origin A, 0 deg is north
it is 30 + (360-343) = 30 + 17 = 47 deg
call Angle in NW quadrant B (quad3) and Angle in NE quadrant(quad1) = C
now
BC^2 = AB^2 + AC^2 -2 AB AC cos 47
but
AB = 2 and AC = 7
so
BC^2 = 4 + 49 - 2 (2)(7)cos 47
solve for BC, now have all 3 sides
then use law of sines to get angles
like
sin C /2 = sin A/BC and Sin B / 7 = same thing
then you have all the sides and all the angles.
Answered by
Yolanda
porkchops are good but this math is better
Answered by
Lort
Ham is awesome
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