Solving triangles: the laws of sines and cosines

Notation and prior facts

In the lesson on trigonometric ratios the sine, the cosine and the tangent of an acute angle were defined as quotients of the sides of a right triangle, and were later extended to every angle by means of the unit circle. A right triangle is therefore solved with these ratios and with the Pythagorean theorem. Now, most of the triangles that appear in geometry and in land surveying have no right angle; the aim of this lesson is to solve them all.

In a triangle with vertices $A$, $B$ and $C$, the same capital letter denotes the vertex and the measure of the interior angle of the triangle at that vertex, and the corresponding lowercase letter denotes the length of the opposite side: $a$ is the length of side $BC$, $b$ that of side $CA$ and $c$ that of side $AB$. The area of the triangle is denoted $S$, and the distance between two points $P$ and $Q$, $\operatorname{d}(P, Q)$. As is customary in solving triangles, angles are expressed in degrees. Now, the trigonometric functions are defined for real numbers, which are measures in radians; by the conversion rule of the lesson on angles and their measure, an angle of $g$ degrees measures $\frac{\pi g}{180}$ radians, and the following convention is adopted: $\sin g^\circ$ denotes $\sin \frac{\pi g}{180}$, and likewise with the other functions; for example, $\cos 60^\circ = \cos \frac{\pi}{3} = \frac{1}{2}$. In the same way, the arccosine, which takes values in $\icc{0}{\pi}$, returns the radian measure of an angle, which is expressed in degrees: the equality $\arccos \left( -\frac{1}{2} \right) = 120^\circ$ means that $\arccos \left( -\frac{1}{2} \right) = \frac{2\pi}{3}$, the radian measure of an angle of $120^\circ$. With this convention, everything that follows holds equally in radians, replacing $180^\circ$ by $\pi$. We admit, as in the lesson on angles and their measure, a fact of elementary geometry that this course does not prove because it does not develop synthetic geometry: the three angles of a triangle add up to two right angles.

It follows that each angle of a triangle lies strictly between $0^\circ$ and $180^\circ$, and that at most one of them is right or obtuse. From the earlier lessons of this unit the following facts are also used:

• The point at distance $r$ from the origin, on the ray that forms the angle $\theta$ with the positive half of the horizontal axis, has the coordinates $\left( r \cos \theta, r \sin \theta \right)$: it is the point of the unit circle corresponding to $\theta$, multiplied by $r$.
• The Pythagorean identity, $\sin^2 \theta + \cos^2 \theta = 1$, and the symmetries of the supplementary angle, $\sin \left( 180^\circ - \theta \right) = \sin \theta$ and $\cos \left( 180^\circ - \theta \right) = -\cos \theta$.
• If $0^\circ < \theta < 180^\circ$, the sine of $\theta$ is positive; the cosine is positive if $\theta$ is acute, zero if it is right and negative if it is obtuse.
• The cosine is strictly decreasing between $0^\circ$ and $180^\circ$ and takes in that interval each value between $-1$ and $1$ exactly once; hence, according to the lesson on the inverse trigonometric functions, an angle of a triangle is determined by its cosine, $\theta = \arccos x$.

The sine, on the other hand, does not determine an angle of a triangle: the angles $\theta$ and $180^\circ - \theta$ have the same sine, and the arcsine function returns only the acute one. This asymmetry between the sine and the cosine will explain, in the penultimate section, why certain data admit two different triangles.