In $\triangle$ABC, let angles A, B, and C be the three interior angles of the triangle, and given that $\frac{sinB}{sinA} = 2\sqrt{3} sinC$, then the range of values for $B + \frac{\pi}{6}$ is_, and the range of values for $\frac{sinC}{sinA} + \frac{sinA}{sinC}$ is_.

This method does not determine the correct range.

FunctionRange[{B + \[Pi]/6, Sin[B]/Sin[A] == 2 Sqrt[3] Sin[C1], 
  A + B + C1 == \[Pi], a/Sin[A] == b/Sin[B] == c/Sin[C1], 
  a^2 + c^2 - b^2 == 2 a  c  Cos[B]}, {a, b, c, A, B, C1}, t]

get the result is:

Since A, B, and C1 are the interior angles of a triangle, each angle ranges within the open interval (0, π). Adding this condition does not yield a more precise or correct answer.

FunctionRange[{B + \[Pi]/6, Sin[B]/Sin[A] == 2 Sqrt[3] Sin[C1], 
  A + B + C1 == \[Pi], a/Sin[A] == b/Sin[B] == c/Sin[C1], 
  a^2 + c^2 - b^2 == 2 a  c  Cos[B], 0 < A < \[Pi], 0 < B < \[Pi], 
  0 < C1 < \[Pi]}, {a, b, c, A, B, C1}, t]

This code get the result.

How can I obtain this final precise result as follows:

In $\triangle$ABC, let angles A, B, and C be the three interior angles of the triangle, and given that $\frac{\sin B}{\sin A} = 2\sqrt{3} \sin C$, then the range of values for $B + \frac{\pi}{6}$ is _ , and the range of values for $\frac{\sin C}{\sin A} + \frac{\sin A}{\sin C}$ is _.

This method does not determine the correct range.

FunctionRange[{B + \[Pi]/6, Sin[B]/Sin[A] == 2 Sqrt[3] Sin[C1], 
  A + B + C1 == \[Pi], a/Sin[A] == b/Sin[B] == c/Sin[C1], 
  a^2 + c^2 - b^2 == 2 a  c  Cos[B]}, {a, b, c, A, B, C1}, t]

get the result is:

Since A, B, and C1 are the interior angles of a triangle, each angle ranges within the open interval (0, π). Adding this condition does not yield a more precise or correct answer.

FunctionRange[{B + \[Pi]/6, Sin[B]/Sin[A] == 2 Sqrt[3] Sin[C1], 
  A + B + C1 == \[Pi], a/Sin[A] == b/Sin[B] == c/Sin[C1], 
  a^2 + c^2 - b^2 == 2 a  c  Cos[B], 0 < A < \[Pi], 0 < B < \[Pi], 
  0 < C1 < \[Pi]}, {a, b, c, A, B, C1}, t]

This code get the result.

How can I obtain this final precise result as follows:

In $\triangle$ABC, let angles A, B, and C be the three interior angles of the triangle, and given that $\frac{sinB}{sinA} = 2\sqrt{3} sinC$, then the range of values for $B + \frac{\pi}{6}$ is_, and the range of values for $\frac{sinC}{sinA} + \frac{sinA}{sinC}$ is_.

This method does not determine the correct range.

FunctionRange[{B + \[Pi]/6, Sin[B]/Sin[A] == 2 Sqrt[3] Sin[C1], 
  A + B + C1 == \[Pi], a/Sin[A] == b/Sin[B] == c/Sin[C1], 
  a^2 + c^2 - b^2 == 2 a  c  Cos[B]}, {a, b, c, A, B, C1}, t]

get the result is:

Since A, B, and C1 are the interior angles of a triangle, each angle ranges within the open interval (0, π). Adding this condition does not yield a more precise or correct answer.

FunctionRange[{B + \[Pi]/6, Sin[B]/Sin[A] == 2 Sqrt[3] Sin[C1], 
  A + B + C1 == \[Pi], a/Sin[A] == b/Sin[B] == c/Sin[C1], 
  a^2 + c^2 - b^2 == 2 a  c  Cos[B], 0 < A < \[Pi], 0 < B < \[Pi], 
  0 < C1 < \[Pi]}, {a, b, c, A, B, C1}, t]

This code get the result.

In $\triangle$ABC, let angles A, B, and C be the three interior angles of the triangle, and given that $\frac{sinB}{sinA} = 2\sqrt{3} sinC$, then the range of values for $B + \frac{\pi}{6}$ is_, and the range of values for $\frac{sinC}{sinA} + \frac{sinA}{sinC}$ is_.

This method does not determine the correct range.

FunctionRange[{B + \[Pi]/6, Sin[B]/Sin[A] == 2 Sqrt[3] Sin[C1], 
  A + B + C1 == \[Pi], a/Sin[A] == b/Sin[B] == c/Sin[C1], 
  a^2 + c^2 - b^2 == 2 a  c  Cos[B]}, {a, b, c, A, B, C1}, t]

get the result is:

Since A, B, and C1 are the interior angles of a triangle, each angle ranges within the open interval (0, π). Adding this condition does not yield a more precise or correct answer.

FunctionRange[{B + \[Pi]/6, Sin[B]/Sin[A] == 2 Sqrt[3] Sin[C1], 
  A + B + C1 == \[Pi], a/Sin[A] == b/Sin[B] == c/Sin[C1], 
  a^2 + c^2 - b^2 == 2 a  c  Cos[B], 0 < A < \[Pi], 0 < B < \[Pi], 
  0 < C1 < \[Pi]}, {a, b, c, A, B, C1}, t]

This code get the result.

How can I obtain this final precise result as follows:

In $\triangle$ABC, let angles A, B, and C be the three interior angles of the triangle, and given that $\frac{sinB}{sinA} = 2\sqrt{3} sinC$, then the range of values for $B + \frac{\pi}{6}$ is_, and the range of values for $\frac{sinC}{sinA} + \frac{sinA}{sinC}$ is_.

This method does not determine the correct range.

FunctionRange[{B + \[Pi]/6, Sin[B]/Sin[A] == 2 Sqrt[3] Sin[C1], 
  A + B + C1 == \[Pi], a/Sin[A] == b/Sin[B] == c/Sin[C1], 
  a^2 + c^2 - b^2 == 2 a  c  Cos[B]}, {a, b, c, A, B, C1}, t]

get the result is:

Since A, B, and C1 are the interior angles of a triangle, each angle ranges within the open interval (0, π). Adding this condition does not yield a more precise or correct answer.

FunctionRange[{B + \[Pi]/6, Sin[B]/Sin[A] == 2 Sqrt[3] Sin[C1], 
  A + B + C1 == \[Pi], a/Sin[A] == b/Sin[B] == c/Sin[C1], 
  a^2 + c^2 - b^2 == 2 a  c  Cos[B], 0 < A < \[Pi], 0 < B < \[Pi], 
  0 < C1 < \[Pi]}, {a, b, c, A, B, C1}, t]

This code has been running without producing a result.

In $\triangle$ABC, let angles A, B, and C be the three interior angles of the triangle, and given that $\frac{sinB}{sinA} = 2\sqrt{3} sinC$, then the range of values for $B + \frac{\pi}{6}$ is_, and the range of values for $\frac{sinC}{sinA} + \frac{sinA}{sinC}$ is_.

This method does not determine the correct range.

FunctionRange[{B + \[Pi]/6, Sin[B]/Sin[A] == 2 Sqrt[3] Sin[C1], 
  A + B + C1 == \[Pi], a/Sin[A] == b/Sin[B] == c/Sin[C1], 
  a^2 + c^2 - b^2 == 2 a  c  Cos[B]}, {a, b, c, A, B, C1}, t]

get the result is:

Since A, B, and C1 are the interior angles of a triangle, each angle ranges within the open interval (0, π). Adding this condition does not yield a more precise or correct answer.

FunctionRange[{B + \[Pi]/6, Sin[B]/Sin[A] == 2 Sqrt[3] Sin[C1], 
  A + B + C1 == \[Pi], a/Sin[A] == b/Sin[B] == c/Sin[C1], 
  a^2 + c^2 - b^2 == 2 a  c  Cos[B], 0 < A < \[Pi], 0 < B < \[Pi], 
  0 < C1 < \[Pi]}, {a, b, c, A, B, C1}, t]

This code get the result.