-4
$\begingroup$
Clear["`*"]
Reduce[{Sin[C] == Sqrt[2] Cos[B], a^2 + b^2 - c^2 == Sqrt[2] a  b, 
  A + B + C == \[Pi], 0 < B < \[Pi], a/Sin[A] == b/Sin[B] == c/Sin[C],
   a^2 == b^2 + c^2 - 2 b  c  Cos[A], 
  b^2 == a^2 + c^2 - 2 a  c  Cos[B], 
  c^2 == b^2 + a^2 - 2  a  b  Cos[C], {a, b, c} > 0}, {A, B, C}, {a, 
  b, c}]

The software kept showing "running" without producing a result when solving this equation. However, the values of these three angles can be computed manually.


An alternative approach can be used to find the values of these three angles.

Clear["`*"]
Reduce[{Sin[C] == Sqrt[2] Cos[B], Cos[C] == 1/Sqrt[2], 
    A + B + C == \[Pi], 0 < {A, B, C} < \[Pi]}, {A, B, C}] // 
  FullSimplify // Reduce

enter image description here

Manually substituting a known condition involving the relationship of sides using the cosine rule to obtain the cosine value of angle C.


But my question is, how can the software be enabled to quickly produce results while keeping the equation in the first code unchanged?


The strange thing is that, while keeping the equation unchanged, the software can solve for each of the three angles individually every time.

enter image description here

 Clear["`*"]
Reduce[{Sin[C] == Sqrt[2] Cos[B], a^2 + b^2 - c^2 == Sqrt[2] a  b, 
    A + B + C == \[Pi], 0 < {A, B, C} < \[Pi], 
    a/Sin[A] == b/Sin[B] == c/Sin[C], 
    a^2 == b^2 + c^2 - 2 b  c  Cos[A], 
    b^2 == a^2 + c^2 - 2 a  c  Cos[B], 
    c^2 == b^2 + a^2 - 2  a  b  Cos[C], {a, b, c} > 0}, {A}, {a, b, c,
     B, C}] // FullSimplify // Reduce

A == (5 \[Pi])/12

 Clear["`*"]
Reduce[{Sin[C] == Sqrt[2] Cos[B], a^2 + b^2 - c^2 == Sqrt[2] a  b, 
    A + B + C == \[Pi], 0 < {A, B, C} < \[Pi], 
    a/Sin[A] == b/Sin[B] == c/Sin[C], 
    a^2 == b^2 + c^2 - 2 b  c  Cos[A], 
    b^2 == a^2 + c^2 - 2 a  c  Cos[B], 
    c^2 == b^2 + a^2 - 2  a  b  Cos[C], {a, b, c} > 0}, {B}, {a, b, c,
     A, C}] // FullSimplify // Reduce

B == \[Pi]/3

Clear["`*"]
Reduce[{Sin[C] == Sqrt[2] Cos[B], a^2 + b^2 - c^2 == Sqrt[2] a  b, 
    A + B + C == \[Pi], 0 < {A, B, C} < \[Pi], 
    a/Sin[A] == b/Sin[B] == c/Sin[C], 
    a^2 == b^2 + c^2 - 2 b  c  Cos[A], 
    b^2 == a^2 + c^2 - 2 a  c  Cos[B], 
    c^2 == b^2 + a^2 - 2  a  b  Cos[C], {a, b, c} > 0}, {C}, {a, b, c,
     A, B}] // FullSimplify // Reduce

C == \[Pi]/4

——————————————————————————————————————

enter image description here

However, when attempting to compute the values of all three angles A, B, and C simultaneously, the software keeps displaying "running" without producing a result.

————————————————————————————————————

Unable to decide whether numeric quantity -Pi + <<2>> - I (Log[1 - I
(Divide[-3 Power[2, <<1>>], (1 - Power[Root[{<<2>>}, {2, 3}], 2])
<<2>>] + <<124>>)] - Log[1 + <<1>>]) is equal to zero. Assuming it is.

enter image description here

$\endgroup$
5
  • 2
    $\begingroup$ This is a typical math problem in plane geometry. I would like to know why you insist on "keeping the equation in the first code unchanged". If you drop this condition, a lot of simple answers will come... $\endgroup$
    – A. Kato
    Commented Jul 9 at 5:30
  • 1
    $\begingroup$ So you're still blindly mimicking the code you've obtained, otherwise you should have tried this basic syntax as the first step: Solve[{ Sin[C] == Sqrt[2] Cos[B], a^2 + b^2 - c^2 == Sqrt[2] a b, A + B + C == \[Pi], 0 < {A, B, C} < \[Pi], a/Sin[A] == b/Sin[B] == c/Sin[C], a^2 == b^2 + c^2 - 2 b c Cos[A], b^2 == a^2 + c^2 - 2 a c Cos[B], c^2 == a^2 + b^2 - 2 a b Cos[C], {a, b, c} > 0}, {A, B, C, a, b, c}] // FullSimplify // AbsoluteTiming (-1) $\endgroup$
    – xzczd
    Commented Jul 9 at 5:47
  • $\begingroup$ @A.Kato This can be solved by conversion as I mentioned above. I just want to know what the cause of this problem is and how to solve it? $\endgroup$
    – csn899
    Commented Jul 9 at 8:36
  • $\begingroup$ @xzczd This also computes the results, but not in the desired format for display. $\endgroup$
    – csn899
    Commented Jul 9 at 10:33
  • 2
    $\begingroup$ So you still don't know how to use Assumptions or Assuming even if they've appeared in the body of your questions so many times: mathematica.stackexchange.com/q/282329/1871 mathematica.stackexchange.com/q/284016/1871 mathematica.stackexchange.com/q/289061/1871 mathematica.stackexchange.com/q/292627/1871 mathematica.stackexchange.com/q/282362/1871 mathematica.stackexchange.com/q/284179/1871 mathematica.stackexchange.com/q/283938/1871 Once again, I don't see your effort. Sadly I cannot cast a second downvote. $\endgroup$
    – xzczd
    Commented Jul 9 at 11:10

1 Answer 1

0
$\begingroup$
Solve[{Sin[C] == Sqrt[2]  Cos[B], a^2 + b^2 - c^2 == Sqrt[2]  a  b, 
    A + B + C == \[Pi], 0 < {A, B, C} < \[Pi], 
    a/Sin[A] == b/Sin[B] == c/Sin[C], 
    a^2 == b^2 + c^2 - 2  b  c  Cos[A], 
    b^2 == a^2 + c^2 - 2  a  c  Cos[B], 
    c^2 == a^2 + b^2 - 2  a  b  Cos[C], {a, b, c} > 0}, {A, B, C, a, 
    b, c}] // FullSimplify // Simplify[#, a > 0] &
Solve[{Sin[C] == Sqrt[2]  Cos[B], a^2 + b^2 - c^2 == Sqrt[2]  a  b, 
    A + B + C == \[Pi], 0 < {A, B, C} < \[Pi], 
    a/Sin[A] == b/Sin[B] == c/Sin[C], 
    a^2 == b^2 + c^2 - 2  b  c  Cos[A], 
    b^2 == a^2 + c^2 - 2  a  c  Cos[B], 
    c^2 == a^2 + b^2 - 2  a  b  Cos[C], {a, b, c} > 0}, {A, B, C, a, 
    b, c}] // FullSimplify // Assuming[a > 0, Refine[#]] &
$\endgroup$

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