Fix the notation

In a non-degenerate triangle, the angles $A,B,C$ are opposite the sides $a,b,c$, respectively. All sides are positive and $A+B+C=\pi$.

Demonstration of the law of cosines

Place vertex A at the origin, B at $(c,0)$ and C at $(b\cos A,b\sin A)$. The distance between B and C is $a$. By Pythagoras’ theorem applied to coordinate differences:

$$\begin{aligned} a^2&=(b\cos A-c)^2+(b\sin A)^2\\ &=b^2(\cos^2A+\sin^2A)+c^2-2bc\cos A\\ &=b^2+c^2-2bc\cos A. \end{aligned}$$

The argument also applies to obtuse angles. If $A=\pi/2$, the cosine term is zero and we find Pythagoras again. Permuting the letters gives the other two cosine laws.

Area and demonstration of the law of sines

The height relative to $c$ measures $b\sin A$, even when the foot falls onto the base extension. The area $K$ is:

$$K=\frac{bc\sin A}{2}=\frac{ac\sin B}{2}=\frac{ab\sin C}{2}.$$

Equating the three expressions and dividing by $abc/2>0$ gives:

$$\frac a{\sin A}=\frac b{\sin B}=\frac c{\sin C}.$$

If $R$ is the radius of the circumscribed circle, a chord subtending the inscribed angle $A$ satisfies $a=2R\sin A$. For high $A$, this comes from dividing the triangle formed by the string and two radii in half; for $A$ obtuse, use the supplement, of the same sine. In the straight case, the chord is a diameter. Therefore:

$$\frac a{\sin A}=\frac b{\sin B}=\frac c{\sin C}=2R,$$

$$K=\frac{abc}{4R}.$$

Projection relations

The displacements about the direction of one side add up, with sign. Projecting the sides $c$ and $b$ onto the straight line of $a$:

$$a=b\cos C+c\cos B.$$

If one of the angles is obtuse, its projection is negative because its cosine is negative. Similarly, $b=a\cos C+c\cos A$ and $c=a\cos B+b\cos A$.

Law of tangents

By the law of sines, for $a\ne b$:

$$\frac{a+b}{a-b}=\frac{\sin A+\sin B}{\sin A-\sin B}.$$

Transforming numerator and denominator into products and simplifying non-zero factors:

$$\frac{a+b}{a-b}= \frac{\tan((A+B)/2)}{\tan((A-B)/2)}.$$

The other relations are obtained by exchanging the pairs of sides and angles. If $a=b$, the expression with denominator $a-b$ does not exist; the triangle must be treated directly as isosceles.

Worked example

If $b=5$, $c=7$ and $A=60°$:

$$a=\sqrt{25+49-70\cdot\frac12}=\sqrt{39},$$

$$K=\frac{5\cdot7\cdot\sin60°}{2}=\frac{35\sqrt3}{4}.$$

Lengths and area remain exact, without rounding radicals.

In everyday life

  • Navigation. Two ships at 8 km and 5 km from the port, on routes at $60^\circ$, are 7 km from each other (law of cosines).
  • Land. The area of ​​a triangular lot extends from two sides and the angle between them: $\frac{1}{2}ab\operatorname{sen} C$.
  • Topography. With a base and two angles, the law of sines gives distances to inaccessible points.
Law of cosines in navigation: the distance between two ships.

Exercises with resolution

1. Two sides measure 6 and 8 cm and the angle between them is $60°$. Calculate the third side.

Check resolution 1

$a^2=36+64-96\cdot1/2=52$, then $a=2\sqrt{13}$ cm.

2. If $a=10$ cm and the opposite angle is $30°$, calculate the circumscribed radius.

Check resolution 2

$2R=a/\sin A=10/(1/2)=20$ cm, then $R=10$ cm.

In the interactive book

The same subjects, with all the demonstrations and the complete list of exercises, are in the interactive Trigonometry book:

This class is a shorter introduction to the topic; In the book, the theory is accompanied by solved and proposed exercises from the chapter.

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