Solve these unkowns x and y using these 2 simultaneous equations
1 vue (au cours des 30 derniers jours)
Afficher commentaires plus anciens
nathalie
le 29 Oct 2023
Commenté : Walter Roberson
le 29 Oct 2023
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
0 commentaires
Réponse acceptée
Walter Roberson
le 29 Oct 2023
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 commentaires
Walter Roberson
le 29 Oct 2023
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)
Voir également
Catégories
En savoir plus sur Numbers and Precision dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!