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In the response to a related question here is given a formula to compute one of the angles not given for an SAS triangle:

$$\tan(A) = \frac{\sin(C)}{\frac{b}{a}-\cos(C)}$$

where $a,b,C$ are the given sides and angle, and $A$ (or $A+\pi$ when $tan(A)<0$) is the angle opposite of side $a$.

Is there a name for this formula? It is similar to the law of tangents (also referenced there) as a method to compute an angle from the given sides and angle without needing to calculate the length of the 3rd side first. A difference is that this formula can generate a negative angle instead of an obtuse angle (in which case it indicates that one must add $\pi$). Here, I refer to it as the "tangent law" as distinguished from the "law of tangents".

smichr
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  • Wouldn't that allow us to know that A=90°? – Gwen Apr 26 '24 at 05:14
  • Yes. And the angle you calculate will be off as demonstrated at the end of the answer here. – smichr Apr 26 '24 at 05:57
  • Gwen, thinking more about this and having edited the answer, your comment may not make sense without context. If $b/a\approx\cos(C)$ then the angle we are computing is almost in a right triangle. This can happen for an isosceles triangle with a very small angle or for an arbitrarily shaped right triangle. In case of the former $A$ is just $(180^o-C)/2$. And in case of the latter we can calculate angle $B$, replacing $b/a$ with $a/b$ and removing the condition of near cancellation, but I haven't convinced myself whether this needs to be done. – smichr Apr 26 '24 at 20:22

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This equation may be obtained by replacing $A+B$ in the tangent law with $\pi - C$ and then expanding the tangents of sums, solving for $tan(A)$ and rewriting $tan(C)$ as $sin(C)/cos(C)$ on the rhs (and simplifying). e.g. here is a demonstration using SymPy:

>>> from sympy.abc import *
>>> from sympy import *
>>> lot = (a - b)/(a + b) - tan(A/2 - B/2)/tan(A/2 + B/2)
>>> solve(lot.subs(B,pi-A-C).expand(trig=True).rewrite(tan),tan(A))
[2*a*tan(C/2)/(a*tan(C/2)**2 - a + b*tan(C/2)**2 + b)]
>>> _[0].rewrite(cos)
2*a*cos(C/2 - pi/2)/((-a + a*cos(C/2 - pi/2)**2/cos(C/2)**2 + b + b*cos(C/2 - pi/2)**2/cos(C/2)**2)*cos(C/2))
>>> expand(_,trig=True).simplify()
a*sin(C)/(-a*cos(C) + b)
>>> (numer(_)/a)/expand(denom(_)/a)
sin(C)/(-cos(C) + b/a)  <--- the equation presented in the OP

The OP is not a new equation, per se, just a re-arrangement of the law of tangents to make it more tractable. It can also be derived from the law of sines by an easier derivation as demonstrated here.

smichr
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