1
$\begingroup$

I am trying to solve the following problem:

Question :In a Triangle,Sin∠A:Sin∠B:Sin∠C = ∠A:∠B:∠C,what's the specific value of ∠A,∠B,∠C? [How to prove A=B=C=π/3 is the only answer]

This is what I have tried so far:

{NSolve[Sin[x]/x == Sin[y]/y == Sin[z]/z && 0 < x < Pi && 
 0 < y < Pi && 0 < z < Pi && x + y + z = Pi, {x, y, z}]}
$\endgroup$
9
  • $\begingroup$ How did you get that equation? What is x and y? $\endgroup$ Apr 10 '17 at 13:21
  • $\begingroup$ Do you know the law of sines, and what the sum of the angles of a triangle is? $\endgroup$
    – J. M.'s torpor
    Apr 10 '17 at 13:21
  • $\begingroup$ @Anjan Kumar Sorry, now I have changed the code $\endgroup$
    – CcccB
    Apr 10 '17 at 13:41
  • $\begingroup$ @J. M.♦ Sorry,now I have changed the code. $\endgroup$
    – CcccB
    Apr 10 '17 at 13:43
  • $\begingroup$ @CcccB Try this: FindRoot[{Sin[x]/x == Sin[y]/y, Sin[y]/y == Sin[z]/z, x + y + z == Pi}, {x, 1}, {y, 1}, {z, 1}] $\endgroup$ Apr 10 '17 at 13:59
2
$\begingroup$

The statement

Sin∠A:Sin∠B:Sin∠C = ∠A:∠B:∠C

means that

$$\frac{∠A}{∠B}=\frac{\sin∠A}{\sin∠B}~~~,~~~\frac{∠B}{∠C}=\frac{\sin∠B}{\sin∠C}~~~,~~~\frac{∠A}{∠C}=\frac{\sin∠A}{\sin∠C}$$

This can be rearranged to

$$\frac{\sin∠B}{∠B}=\frac{\sin∠A}{∠A}~~~,~~~\frac{\sin∠C}{∠C}=\frac{\sin∠B}{∠B}~~~,~~~\frac{\sin∠C}{∠C}=\frac{\sin∠A}{∠A}$$

Treating ∠A,∠B,∠C as independent continuous variables, we see that all three of them have to be constant, since left hand side and right hand side of all above equations depend on disjoint variables only. Additionally, due to the complete symmetry of how ∠A,∠B,∠C enter the problem, all three of them will have the same value $\alpha$. In a triangle we have

$$∠A+∠B+∠C=3\alpha=\pi$$

Which is sufficient to conclude $\alpha=\frac{\pi}{3}$.

UPDATE

Additionally, and maybe more appropriately for Mathematica use, once we eliminate i.e. ∠C as ∠C=Pi-∠A-∠B, we can explicitly check that within the appropriate range of angles there exists only exactly one solution:

ContourPlot[{Sin[x]/x - Sin[\[Pi] - x - y]/(\[Pi] - x - y) == 0, 
    Sin[y]/y - Sin[\[Pi] - x - y]/(\[Pi] - x - y) == 0, , Sin[x]/x - Sin[y]/y == 0}, 
    {x, 0, \[Pi]}, {y, 0, \[Pi]}, WorkingPrecision -> 30]// Quiet

enter image description here

Which must be the one we found.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.