Solve these unkowns x and y using these 2 simultaneous equations

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nathalie 2023-10-29

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2040216-solve-these-unkowns-x-and-y-using-these-2-simultaneous-equations

评论： Walter Roberson 2023-10-29

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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Answer 1

Walter Roberson 2023-10-29

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2040216-solve-these-unkowns-x-and-y-using-these-2-simultaneous-equations#answer_1342996

在 MATLAB Online 中打开

syms m3R3l3 m4R4l4

Eq1 = 2760 * sind(sym(200)) + m3R3l3 * sind(sym(107)) + m4R4l4 * sind(sym(307)) == 0

Eq1 =

Eq2 = 2760 * cosd(sym(200)) + m3R3l3 * cosd(sym(107)) + m4R4l4 * cosd(sym(307)) == 0

Eq2 =

sol = solve([Eq1, Eq2])

sol = struct with fields:

m3R3l3: (2760*(cos(pi/9)*sin((53*pi)/180) + cos((53*pi)/180)*sin(pi/9)))/(cos((53*pi)/180)*sin((73*pi)/180) - cos((73*pi)/180)*sin((53*pi)/180)) m4R4l4: (2760*(cos(pi/9)*sin((73*pi)/180) + cos((73*pi)/180)*sin(pi/9)))/(cos((53*pi)/180)*sin((73*pi)/180) - cos((73*pi)/180)*sin((53*pi)/180))

simplify(sol.m3R3l3)

ans =

simplify(sol.m4R4l4)

ans =

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nathalie 2023-10-29

Yes there any way to solve this final answer to get normal numbers? On matlab?

Walter Roberson 2023-10-29

在 MATLAB Online 中打开

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

Eq1 =

Eq2 = 2760 * cosd(sym(200)) + m3R3l3 * cosd(sym(107)) + m4R4l4 * cosd(sym(307)) == 0

Eq2 =

sol = solve([Eq1, Eq2])

sol = struct with fields:

m3R3l3: (2760*(cos(pi/9)*sin((53*pi)/180) + cos((53*pi)/180)*sin(pi/9)))/(cos((53*pi)/180)*sin((73*pi)/180) - cos((73*pi)/180)*sin((53*pi)/180)) m4R4l4: (2760*(cos(pi/9)*sin((73*pi)/180) + cos((73*pi)/180)*sin(pi/9)))/(cos((53*pi)/180)*sin((73*pi)/180) - cos((73*pi)/180)*sin((53*pi)/180))

simplify(sol.m3R3l3, 'steps', 1000)

ans =

simplify(sol.m4R4l4, 'steps', 1000)

ans =

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