A triangle has one side that measures 12 inches and another side that measures 33 inches. Which are possible side lengths of the
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This can be solved a couple of ways. One way is to use the Pythagorean theorem to write equations for the magnitude from the components of the forces. That is what was done in the graph here.
Another way is to use the Law of Cosines, which lets you make direct use of the angle between the vectors.
.. 13 = a^2 +b^2 -2ab*cos(90°)
.. 19 = a^2 +b^2 -2ab*cos(120°)
Subtracting the first equation from the second, we have
.. 6 = -2ab*cos(120°)
.. ab = 6
Substituting this into the first equation, we have
.. 13 = a^2 +(6/a)^2
.. a^4 -13a^2 +36 = 0
.. (a^2 -9)(a^2 -4) = 0
.. a = ±3 or ±2
The magnitudes of the two forces are 2N and 3N, in no particular order.
Another way is to use the Law of Cosines, which lets you make direct use of the angle between the vectors.
.. 13 = a^2 +b^2 -2ab*cos(90°)
.. 19 = a^2 +b^2 -2ab*cos(120°)
Subtracting the first equation from the second, we have
.. 6 = -2ab*cos(120°)
.. ab = 6
Substituting this into the first equation, we have
.. 13 = a^2 +(6/a)^2
.. a^4 -13a^2 +36 = 0
.. (a^2 -9)(a^2 -4) = 0
.. a = ±3 or ±2
The magnitudes of the two forces are 2N and 3N, in no particular order.