Four theorems complete the trigonometric properties of triangles: one decomposes each side into the projections of the other two, two give the area and the last, the law of tangents, relates sums and differences of sides to sums and differences of angles.

The story

The law of tangents was published by the Danish Thomas Fincke in 1583, in the same book in which he named the tangent (see chapter VIII), and François Viète also used it at the same time. It became very practical in the era of logarithm tables: as it only involves quotients and products, it allows you to solve a triangle knowing two sides and the angle between them without adding squares, which the law of cosines would require. With logarithms, each multiplication or division became an addition or subtraction of numbers taken from the table.

With logarithms, multiplying is adding: it is the principle of Napier’s tables and the slide rule, which made the law of tangents so practical.
  1. He is born in Merchiston Tower, near Edinburgh, the son of a 16-year-old nobleman.
  2. He enters Scotland’s oldest university at 13 and leaves without a degree.
  3. He studies abroad, probably in France or the Low Countries. The records do not say where; the point on the map is only indicative.
  4. He lives in the castle he built on the family lands. He writes on the Apocalypse and begins the calculations of the logarithms.
  5. He signs a contract with Robert Logan to find treasure hidden in the castle, in exchange for a third of it.
  6. Laird of Merchiston. He publishes the logarithms (1614), receives Briggs (1615), publishes the Rabdologia and dies in 1617.
  7. At Gresham College Henry Briggs reads the book and sets off for Edinburgh to meet the author.
  8. Kepler sees the tables, writes to the baron without knowing he has died and uses logarithms in the Rudolphine Tables.
Napier’s places in Scotland, and the journeys of his tables to London and to Kepler. Map: © OpenStreetMap contributors

Sources: Fincke (MacTutor) · Viète (MacTutor) · Napier (MacTutor).

In everyday life

  • Land. A triangular piece of land with sides of 40 m and 30 m forming $60^\circ$ has an area of ​​$\frac{1}{2} \cdot 40 \cdot 30 \operatorname{sen} 60^\circ \approx 519{,}6$ m².
  • Irregular areas. Surveyors and map programs divide any polygon into triangles and add the areas; this is how the area of ​​a lot or a farm is calculated.
  • Works and renovations. To buy flooring or grass for a triangular flowerbed, simply measure two sides and the angle between them.
The area of ​​a triangular plot of land from two sides and the angle between them.

Theory

Projection theorem

TeoremaProjection theorem

$$a = b\cos\hat{C} + c\cos\hat{B} \qquad b = a\cos\hat{C} + c\cos\hat{A} \qquad c = a\cos\hat{B} + b\cos\hat{A}$$
View the demo

Let’s try the first; the others are analogous. Let $\overline{AD}$ be the height relative to $\overline{BC}$.

1st case: $\hat{B} < 90^\circ$ and $\hat{C} < 90^\circ$. $D$ is between $B$ and $C$. In the right triangle $ABD$, $BD = c\cos\hat{B}$; in the right triangle $ADC$, $DC = b\cos\hat{C}$. Then $a = BD + DC = c\cos\hat{B} + b\cos\hat{C}$.

2nd case: $\hat{B}$ obtuse (the case of $\hat{C}$ obtuse is symmetric). $D$ falls in the extension of $\overline{CB}$, beyond $B$; $BD = c\cos(180^\circ - \hat{B}) = -c\cos\hat{B}$ and $DC = b\cos\hat{C}$. Then $a = DC - DB = b\cos\hat{C} + c\cos\hat{B}$.

3rd case: $\hat{B}$ or $\hat{C}$ straight. $D$ coincides with the vertex of the right angle and the formula continues to be valid, as $\cos 90^\circ = 0$.

Area as a function of two sides and the angle between them

TeoremaTriangle area

In any triangle, the area is equal to the semiproduct of two sides multiplied by the sine of the angle they form:

$$S = \frac{1}{2}bc\operatorname{sen}\hat{A} = \frac{1}{2}ac\operatorname{sen}\hat{B} = \frac{1}{2}ab\operatorname{sen}\hat{C}$$
View the demo

Let $\overline{BD}$ be the height relative to the side $\overline{AC}$. If $\hat{A} < 90^\circ$, in the right triangle $ADB$ we have $DB = c\operatorname{sen}\hat{A}$; if $90^\circ < \hat{A} < 180^\circ$, $D$ is in the extension of $\overline{CA}$ and $DB = c\operatorname{sen}(180^\circ - \hat{A}) = c\operatorname{sen}\hat{A}$. In both cases (and also if $\hat{A} = 90^\circ$),

$$S = \frac{AC \cdot DB}{2} = \frac{b \cdot c\operatorname{sen}\hat{A}}{2}.$$

The other two formulas are proven in the same way.

With two fixed sides, the area $S = \frac{1}{2}bc\operatorname{sen}\hat{A}$ is maximum when the angle between them is right (here, $\frac{1}{2}bc = 1$).

Area as a function of the radius of the circumscribed circle

TeoremaArea and circumscribed radius

In any triangle, the area is equal to the product of the three sides divided by four times the radius of the circumscribed circle:

$$S = \frac{abc}{4R}$$
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By the law of sines, $\operatorname{sen}\hat{A} = \dfrac{a}{2R}$. Replacing $S = \dfrac{bc}{2}\operatorname{sen}\hat{A}$ comes $S = \dfrac{bc}{2} \cdot \dfrac{a}{2R} = \dfrac{abc}{4R}$.

Law of tangents

TeoremaLaw of tangents

In any non-isosceles triangle, the following apply:

$$\frac{a + b}{a - b} = \frac{\operatorname{tg}\frac{\hat{A} + \hat{B}}{2}}{\operatorname{tg}\frac{\hat{A} - \hat{B}}{2}}$$$$\frac{a + c}{a - c} = \frac{\operatorname{tg}\frac{\hat{A} + \hat{C}}{2}}{\operatorname{tg}\frac{\hat{A} - \hat{C}}{2}}$$$$\frac{b + c}{b - c} = \frac{\operatorname{tg}\frac{\hat{B} + \hat{C}}{2}}{\operatorname{tg}\frac{\hat{B} - \hat{C}}{2}}$$

(In formulas, the sides of the differences must be unequal, and none of the angles involved can be right for the tangents to exist).

View the demo

From the law of sines, $\dfrac{a}{b} = \dfrac{\operatorname{sen}\hat{A}}{\operatorname{sen}\hat{B}}$. By the property of proportions (“the sum is to the difference…”):

$$\frac{a + b}{a - b} = \frac{\operatorname{sen}\hat{A} + \operatorname{sen}\hat{B}}{\operatorname{sen}\hat{A} - \operatorname{sen}\hat{B}} = \frac{2\operatorname{sen}\frac{\hat{A} + \hat{B}}{2}\cos\frac{\hat{A} - \hat{B}}{2}}{2\operatorname{sen}\frac{\hat{A} - \hat{B}}{2}\cos\frac{\hat{A} + \hat{B}}{2}} = \frac{\operatorname{tg}\frac{\hat{A} + \hat{B}}{2}}{\operatorname{tg}\frac{\hat{A} - \hat{B}}{2}}.$$

The other two are proved in an analogous way.

The exercises in this section are numbers 393 to 399 from chapter list.