2.3 The Law of Tangents

We have shown how to solve a triangle in all four cases discussed at the beginning of this chapter. An alternative to the Law of Cosines for Case 3 (two sides and the included angle) is the Law of Tangents:

Theorem 2.3

Law of Tangents: If a triangle has sides of lengths \(a\), \(b\), and \(c\) opposite the angles \(A\), \(B\), and \(C\), respectively, then

\begin{align} \frac{a-b}{a+b} ~&=~ \frac{\tan\;\frac{1}{2}(A-B)}{\tan\;\frac{1}{2}(A+B)} ~,\tag{2.17}\\ \frac{b-c}{b+c} ~&=~ \frac{\tan\;\frac{1}{2}(B-C)}{\tan\;\frac{1}{2}(B+C)} ~,\tag{2.18}\\ \frac{c-a}{c+a} ~&=~ \frac{\tan\;\frac{1}{2}(C-A)}{\tan\;\frac{1}{2}(C+A)} ~. \tag{2.19}\end{align}

Note that since \(\tan\;(-\theta) = -\tan\;\theta\) for any angle \(\theta\), we can switch the order of the letters in each of the above formulas. For example, we can rewrite formula (2.17) as

\begin{equation} \frac{b-a}{b+a}~=~\frac{\tan\;\frac{1}{2}(B-A)}{\tan\;\frac{1}{2}(B+A)}~, \tag{2.20}\end{equation}

and similarly for the other formulas. If \(a > b\), then it is usually more convenient to use formula (2.17), while formula (2.20) is more convenient when \(b > a\).

Example 2.10

[r]tikzpicture[scale=0.8,every node/.style=font=] [fillcolor] (0,0) -- (1.2,1.2) -- (3,0) -- cycle; [linecolor,line width=1.5pt] (0,0) -- (1.2,1.2) node[black,midway,above left] b=3 -- (3,0) node[black,midway,above right] a=5 -- cycle; [below] at (1.5,0) c; [below left] at (0,0) A; [below right] at (3,0) B; [above] at (1.2,1.2) C=96 ; tikzpicture Case 3: Two sides and the included angle.
Solve the triangle \(\triangle\,ABC\) given \(a =5\), \(b = 3\), and \(C = 96\Degrees\).

Solution: \(A + B + C = 180\Degrees\), so \(A + B = 180\Degrees - C = 180\Degrees - 96\Degrees = 84\Degrees\). Thus, by the Law of Tangents,

\begin{align*} \frac{a-b}{a+b} ~=~ \frac{\tan\;\frac{1}{2}(A-B)}{\tan\;\frac{1}{2}(A+B)} \quad&\Rightarrow\quad \frac{5-3}{5+3} ~=~ \frac{\tan\;\frac{1}{2}(A-B)}{\tan\;\frac{1}{2}(84\Degrees)}\\ &\Rightarrow\quad \tan\;\tfrac{1}{2}(A-B) ~=~ \tfrac{2}{8}\tan\;42\Degrees ~=~ 0.2251\\ &\Rightarrow\quad \tfrac{1}{2}(A-B) ~=~ 12.7\Degrees \quad\Rightarrow\quad A-B ~=~ 25.4\Degrees ~. \end{align*}

We now have two equations involving \(A\) and \(B\), which we can solve by adding the equations:

\begin{alignat*}{3} A &- B &&=\; 25.4\Degrees\\ A &+ B &&=\; 84\Degrees\phantom{4\Degrees}\\[-2mm] --&--&&----\\[-2mm] 2A &\phantom{+} &&=\; 109.4\Degrees \quad\Rightarrow\quad \boxed{A = 54.7\Degrees} \quad\Rightarrow\quad B ~=~ 84\Degrees - 54.7\Degrees \quad\Rightarrow\quad \boxed{B = 29.3\Degrees} \end{alignat*}

We can find the remaining side \(c\) by using the Law of Sines:

\begin{displaymath} c ~=~ \frac{a\;\sin\;C}{\sin\;A} ~=~ \frac{5\;\sin\;96\Degrees}{\sin\;54.7\Degrees} \quad\Rightarrow\quad \boxed{c = 6.09} \end{displaymath}

Note that in any triangle \(\triangle\,ABC\), if \(a = b\) then \(A = B\) (why?), and so both sides of formula (2.17) would be \(0\) (since \(\tan\;0\Degrees = 0\)). This means that the Law of Tangents is of no help in Case 3 when the two known sides are equal. For this reason, and perhaps also because of the somewhat unusual way in which it is used, the Law of Tangents seems to have fallen out of favor in trigonometry books lately. It does not seem to have any advantages over the Law of Cosines, which works even when the sides are equal, requires slightly fewer steps, and is perhaps more straightforward.[1]

Related to the Law of Tangents are Mollweide’s equations:[2]

Mollweide’s equations: For any triangle \(\triangle\,ABC\),

\begin{align} \frac{a-b}{c} ~&=~ \frac{\sin\;\frac{1}{2}(A-B)}{\cos\;\frac{1}{2}C} ~,~\text{and}\tag{2.21}\\ \frac{a+b}{c} ~&=~ \frac{\cos\;\frac{1}{2}(A-B)}{\sin\;\frac{1}{2}C} ~. \tag{2.22}\end{align}

Note that all six parts of a triangle appear in both of Mollweide’s equations. For this reason, either equation can be used to check a solution of a triangle. If both sides of the equation agree (more or less), then we know that the solution is correct.

Example 2.11

Use one of Mollweide’s equations to check the solution of the triangle from Example 2.10.

Solution: Recall that the full solution was \(a=5\), \(b=3\), \(c=6.09\), \(A=54.7\Degrees\), \(B=29.3\Degrees\), and \(C=96\Degrees\). We will check this with equation (2.21):

\begin{align*} \frac{a-b}{c} ~&=~ \frac{\sin\;\frac{1}{2}(A-B)}{\cos\;\frac{1}{2}C}\\ \frac{5-3}{6.09} ~&=~ \frac{\sin\;\frac{1}{2}(54.7\Degrees - 29.3\Degrees)}{\cos\;\frac{1}{2}(96\Degrees)}\\ \frac{2}{6.09} ~&=~ \frac{\sin\;12.7\Degrees}{\cos\;48\Degrees}\\ 0.3284 ~&=~ 0.3285 \quad\text{\ding{51}} \end{align*}

The small difference (\(\approx 0.0001\)) is due to rounding errors from the original solution, so we can conclude that both sides of the equation agree, and hence the solution is correct.

Example 2.12

Can a triangle have the parts \(a=6\), \(b=7\), \(c=9\), \(A=55\Degrees\), \(B=60\Degrees\), and \(C=65\Degrees\;\)?

Solution: Before using Mollweide’s equations, simpler checks are that the angles add up to \(180\Degrees\) and that the smallest and largest sides are opposite the smallest and largest angles, respectively. In this case all those conditions hold. So check with Mollweide’s equation (2.22):

\begin{align*} \frac{a+b}{c} ~&=~ \frac{\cos\;\frac{1}{2}(A-B)}{\sin\;\frac{1}{2}C}\\ \frac{6+7}{9} ~&=~ \frac{\cos\;\frac{1}{2}(55\Degrees - 60\Degrees)}{\sin\;\frac{1}{2}(65\Degrees)}\\ \frac{13}{9} ~&=~ \frac{\cos\;(-2.5\Degrees)}{\sin\;32.5\Degrees}\\ 1.44 ~&=~ 1.86 \quad\text{\ding{55}} \end{align*}

Here the difference is far too large, so we conclude that there is no triangle with these parts.

We will prove the Law of Tangents and Mollweide’s equations in Chapter 3, where we will be able to supply brief analytic proofs.[3]

Exercises

For Exercises 1-3, use the Law of Tangents to solve the triangle \(\triangle\,ABC\).

3

  1. \(a = 12\), \(b = 8\), \(C = 60\Degrees\)

  2. \(A = 30\Degrees\), \(b = 4\), \(c = 6\)

  3. \(a = 7\), \(B = 60\Degrees\), \(c = 9\)

For Exercises 4-6, check if it is possible for a triangle to have the given parts.

  1. \(a=5\), \(b=7\), \(c=10\), \(A=27.7\Degrees\), \(B=40.5\Degrees\), \(C=111.8\Degrees\)

  2. \(a=3\), \(b=7\), \(c=9\), \(A=19.2\Degrees\), \(B=68.2\Degrees\), \(C=92.6\Degrees\)

  3. \(a=6\), \(b=9\), \(c=9\), \(A=39\Degrees\), \(B=70.5\Degrees\), \(C=70.5\Degrees\)

  4. Let \(\triangle\,ABC\) be a right triangle with \(C=90\Degrees\). Show that \(\;\tan\;\frac{1}{2}(A-B) =\frac{a-b}{a+b}\,\).

  5. For any triangle \(\triangle\,ABC\), show that \(\;\tan\;\frac{1}{2}(A-B) = \frac{a-b}{a+b}\;\cot\;\frac{1}{2}C\,\).

  6. For any triangle \(\triangle\,ABC\), show that \(\;\tan\;A = \dfrac{a\;\sin\;B}{c - a\;\cos\;B}\,\). (Hint: Draw the altitude from the vertex \(C\) to \(\overline{AB}\).) Notice that this formula provides another way of solving a triangle in Case 3 (two sides and the included angle).

  7. For any triangle \(\triangle\,ABC\), show that \(\;c = b\;\cos\;A + a\;\cos\;B\,\). This is another check of a triangle.

  8. If \(\,b\;\cos\;A = a\;\cos\;B\,\), show that the triangle \(\triangle\,ABC\) is isosceles.

  9. Let \(ABCD\) be a quadrilateral which completely contains its two diagonals. The quadrilateral has eight parts: four sides and four angles. What is the smallest number of parts that you would need to know to solve the quadrilateral? Explain your answer.

  1. Before the advent of electronic calculators, the Law of Tangents was more popular than it is today since it lent itself better than the Law of Cosines to what was known as logarithmic computation. In those days, computations with large numbers were handled by taking logarithms and looking up values in a logarithm table. Ratios (such as in the Law of Tangents and the Law of Sines) could be replaced by differences of logarithms, making computation easier. ↩
  2. Named after the German astronomer and mathematician Karl Mollweide (1774-1825). ↩
  3. There are (complex) geometric proofs of the Law of Tangents and Mollweide’s equations. See pp. 96-98 in P.R. Rider, Plane and Spherical Trigonometry, New York: The Macmillan Company, 1942. ↩