The Law of Tangents
Article objectives
For a general triangle, which may or may not have a right angle, we will again need three pieces of information. The four cases are:
Case 1: One side and two angles
Case 2: Two sides and one opposite angle
Case 3: Two sides and the angle between them
Case 4: Three sides
An alternative to the Law of Cosines for Case 3 (two sides and the included angle) is the Law of Tangents:
Law of Tangents: If a triangle has sides of lengths a, b, and c opposite the angles A, B, and C, respectively, then $$\frac{a-b}{a+b} = \frac{tan\frac{1}{2}(A-B)}{tan\frac{1}{2}(A+B)},\;\;\;\;\; (1)$$ $$\frac{b-c}{b+c} = \frac{tan\frac{1}{2}(B-C)}{tan\frac{1}{2}(B+C)},\;\;\;\;\; (2)$$ $$\frac{c-a}{c+a} = \frac{tan\frac{1}{2}(C-A)}{tan\frac{1}{2}(C+A)},\;\;\;\;\; (3)$$
Note that since tan (−θ) = −tan θ for any angle θ, we can switch the order of the letters in each of the above formulas. For example, we can rewrite formula (1) as
$$ \frac{b−a}{b+a} = \frac{tan \frac{1}{2} (B− A)} {tan \frac{1}{2}(B+ A)}\;\;\;\;\; (4)$$
and similarly for the other formulas. If a > b, then it is usually more convenient to use formula (1), while formula (4) is more convenient when b > a.
Example 1
Case 3: Two sides and the included angle. Solve the triangle △ABC given a = 5, b = 3, and C = 96º.
Solution: A+B+C = 180º, so A+B = 180º −C = 180º −96º = 84º. Thus, by the Law of Tangents,
$$\frac{a-b}{a+b} = \frac{tan\frac{1}{2}(A-B)}{tan\frac{1}{2}(A+B)} ⇒ \frac{5−3}{ 5+3} = \frac{tan \frac{1}{2} (A−B)}{ tan \frac{1 }{2} (84º)}$$ $$⇒ tan \frac{1}{2} (A−B) = \frac{2}{8} tan 42º = 0.2251$$ $$⇒ \frac{1}{ 2} (A−B) = 12.7º ⇒ A−B = 25.4º .$$
We now have two equations involving A and B, which we can solve by adding the equations:
| $$A-B=25.4º$$ |
| $$A+B=84º$$ |
$$2A=109.4º ⇒ \boxed{A = 54.7º} ⇒ B = 84º −54.7º ⇒ \boxed{B = 29.3º}$$
We can find the remaining side c by using the Law of Sines: $$c = \frac{a sin C}{ sin A} = \frac{5 sin 96º}{ sin 54.7º} ⇒ \boxed{c = 6.09}$$
Note that in any triangle △ABC, if a = b then A = B (why?), and so both sides of formula (1) would be 0 (since tan 0º = 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. Related to the Law of Tangents are Mollweide’s equations:
Mollweide's equations For any triangle △ABC,
$$\frac{a−b}{ c} = \frac{sin \frac{1}{ 2}(A−B)}{cos \frac{1}{ 2} C}, and\;\;\;\;\; (5)$$ $$\frac{a+b}{ c} = \frac{cos \frac{1}{ 2}(A−B)}{sin \frac{1}{ 2} C}\;\;\;\;\; (6)$$
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
Use one of Mollweide’s equations to check the solution of the triangle from Example 1.
Solution: Recall that the full solution was a = 5, b = 3, c = 6.09, A = 54.7º, B = 29.3º, and C = 96º. We will check this with equation (5):
$$\frac{a−b}{ c} = \frac{sin \frac{1}{ 2}(A−B)}{cos \frac{1}{ 2} C}$$ $$\frac{5−3}{ 6.09} = \frac{sin \frac{1}{ 2}(54.7º−29.3º)}{cos \frac{1}{ 2} (96º)}$$ $$\frac{2}{ 6.09} = \frac{sin(12.7º)}{cos(48º)}$$ $$0.3284 = 0.3285 ✓$$
The small difference (≈ 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 3
Can a triangle have the parts a = 6, b = 7, c = 9, A = 55º, B = 60º, and C = 65º ?
Solution: Before using Mollweide’s equations, simpler checks are that the angles add up to 180º 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:
$$\frac{a+b}{ c} = \frac{cos \frac{1}{ 2}(A−B)}{sin \frac{1}{ 2} C}$$ $$\frac{6+7}{ 9} = \frac{cos \frac{1}{ 2}( 55º−60º)}{sin \frac{1}{ 2} (65º)}$$ $$\frac{13}{ 9} = \frac{cos (−2.5º)}{sin (32.5º)}$$ $$1.44 = 1.86 ✗$$
Here the difference is far too large, so we conclude that there is no triangle with these parts.