Solving the Law of Tangents for an Angle

Date: 12/18/2004 at 01:19:38
From: Jeff
Subject: Law of Tangents

For (a - b)/(a + b) = tan (A/2 - B/2) / tan (A/2 + B/2), given
a = 3, b = 2, and A = 40, solve for B using only the Law of Tangents.

This question can easily be answered using the Law of Sines.  But I 
can not answer it using the above info only.  It must be done using 
the Law of Tangents only and we do not know what c is and do not need 
it.  I was tutoring a student and her teacher gave this to the class
as a portfolio problem.  Turns out her teacher could not solve it, nor
could I, and the problem was then dropped.  Being a math major, the
problem interested me and it was the first I had heard of this Law. 

If given A and B then a or b can be found with ease when you know one 
of a or b and are solving for the other.  The problem I am having is 
solving the Law of Tangents for B.

I have used various identities to try to solve this and I always get 
a longer and more complicated equation.  I would like to know what B 
equals in general, in terms of a, b, and A.  I could show you the work 
I have done, but it is long and leads to a dead end or like I stated, 
a larger equation.

Date: 12/18/2004 at 09:03:45
From: Doctor Pete
Subject: Re: Law of Tangents

Hi Jeff,

Let us begin with the following useful trigonometric identity:  For
any angle t, we have

     sin[2t] = 2 sin[t]cos[t] = 2 tan[t]/sec[t]^2 = 2 tan[t]/(1 +
tan[t]^2).

Now we are given

     1/5 = tan[20 - B/2]/tan[20 + B/2].

Let a = tan[20], and x = tan[B/2].  It follows from the tangent angle
addition formula that

     1/5 = (a - x)(1 - ax)/((1 + ax)(a + x)),

whereupon expanding, cross-multiplying, and collecting like terms on
one side, we obtain

     4ax^2 - 6(a^2 + 1)x + 4a = 0,

and since 4a is nonzero, we get the quadratic

     0 = x^2 - 3((a^2 + 1)/(2a)) x + 1
       = x^2 - 3((tan[20]^2 + 1)/(2 tan[20])) x + 1
       = x^2 - 3 csc[40] x + 1,

the last equality due to the previously mentioned identity.  Therefore

     x^2 + 1 = 3 csc[40] x,

or equivalently,

     (2/3) sin[40] = 2x/(x^2 + 1) = 2 tan[B/2]/(tan[B/2]^2 + 1) = sin[B],

again the last equality being the consequence of the same identity. 
Therefore,

     B = arcsin[(2/3) sin[40]],

which is the same as what one would have obtained using the Law of Sines.

-*-*-*-

Indeed, then, our solution leads us to suspect that the identity

     sin[2t] = 2 tan[t]/(tan[t]^2 + 1)

is intimately related to the Law of Tangents.  From the Law of Sines,

     Sin[A]/a = Sin[B]/b,

or

     2b tan[A/2](tan[B/2]^2 + 1) = 2a tan[B/2](tan[A/2]^2 + 1).

For simplicity, let x = tan[A/2], y = tan[B/2].  We then have

     2b x(y^2 + 1) = 2a y(x^2 + 1).

Now add by(x^2 + 1) + ax(y^2 + 1) to both sides to obtain

  (a+b)x(y^2+1) + b(x(y^2+1) + y(x^2+1)) = (a+b)y(x^2+1) + a(y(x^2+1)
+ x(y^2+1)).

Rearranging, we get

     (a+b)(y(x^2+1) - x(y^2+1)) = (a-b)(x(y^2+1) + y(x^2+1)),

or

     (a+b)(x-y)(1-xy) = (a-b)(x+y)(1+xy),

which in turn gives

     (a+b)(x-y)/(1+xy) = (a-b)(x+y)/(1-xy).

Substituting back the values of x and y, we recognize the tangent
angle addition formula to find

     (a+b)tan[(A-B)/2] = (a-b)tan[(A+B)/2],

and the Law of Tangents follows immediately.  So in fact this Law is
actually equivalent to the Law of Sines.

A little thought, then, reveals that were you to solve the Law of
Tangents for the angle B in terms of a, b, and A, you would eventually
obtain

     B = arcsin[(b/a) sin[A]],

which is the same solution for the Law of Sines.  Essentially, you
would be proceeding backwards through the above proof.

I hope my answer has been interesting and satisfactory--I could very
well have gone through the proof demonstrating the equivalence of the
two Laws, and then subsequently used it to solve the question, but it
seems more enlightening to start from the simpler, concrete case.  

You may wonder how I "knew" to start with the very first identity, or
to add some seemingly inspired quantity to both sides of the equation
during one step of the proof.  The answer, of course, is that these
things were found only after working through the problem in the
reverse direction.

- Doctor Pete, The Math Forum
  http://mathforum.org/dr.math/