Solve these unkowns x and y using these 2 simultaneous equations
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Eq1= 2760 * sin (200) + m3R3L3 * sin (107) + m4R4L4 * sin (307) = 0
Eq2= 2760 * cos (200) + m3R3l3 * cos (107) + m4R4L4 * cos(307) = 0
I want to get m3r3l3 and m3r4l4 we can consider m3r3l3 as x and m4r4l4 as y
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Walter Roberson
2023 年 10 月 29 日
syms m3R3l3 m4R4l4
Eq1 = 2760 * sind(sym(200)) + m3R3l3 * sind(sym(107)) + m4R4l4 * sind(sym(307)) == 0
Eq2 = 2760 * cosd(sym(200)) + m3R3l3 * cosd(sym(107)) + m4R4l4 * cosd(sym(307)) == 0
sol = solve([Eq1, Eq2])
simplify(sol.m3R3l3)
simplify(sol.m4R4l4)
2 件のコメント
Walter Roberson
2023 年 10 月 29 日
No, π is transcendental. It is mathematically impossible to express it in terms of a finite series of "algebraic numbers". It is not the root of any finite polynomial with rational coefficients. π is one of the most abnormal real numbers that exist.
However you can get a more compact answer than the above:
syms m3R3l3 m4R4l4
Eq1 = 2760 * sind(sym(200)) + m3R3l3 * sind(sym(107)) + m4R4l4 * sind(sym(307)) == 0
Eq2 = 2760 * cosd(sym(200)) + m3R3l3 * cosd(sym(107)) + m4R4l4 * cosd(sym(307)) == 0
sol = solve([Eq1, Eq2])
simplify(sol.m3R3l3, 'steps', 1000)
simplify(sol.m4R4l4, 'steps', 1000)
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