A Triangle with Three Right Angles, math that surprises you lesson

On a flat sheet of paper the three angles of a triangle always add up to 180 degrees, so a triangle with three right angles sounds impossible. On a globe it is easy to make one. Walk from the North Pole straight down to the equator, turn left by a right angle and walk a quarter of the way around the equator, then turn left by a right angle again and walk straight back up to the pole, where your two paths meet at a right angle too. This page explains why that shape is a genuine triangle in the geometry of the sphere, how its angles are measured, and how the extra 90 degrees is tied exactly to its area.

Key terms

Spherical triangle

Definition Take three points on a sphere that do not all lie on one great circle. No two of them are then opposite ends of a diameter, so each pair lies on exactly one great circle, and the two points split it into a shorter and a longer arc. Join each pair by the shorter arc. These three arcs are the sides, they meet only at the three points, which are the vertices, and together they cut the sphere into two regions. The spherical triangle is the smaller region, together with its sides. Its angle at a vertex is the angle between the two sides that meet there, measured between the directions in which they leave the vertex, that is, in the flat plane that touches the sphere at that point. Every angle of a spherical triangle lies strictly between $0^\circ$ and $180^\circ$ , and the angle sum is always more than $180^\circ$ and less than $540^\circ$ . The rule of $180^\circ$ is a theorem about triangles in a flat plane and does not carry over. The triangle with three right angles has the North Pole $N$ and two points $A$ and $B$ of the equator, $90^\circ$ apart in longitude, as its vertices. Its sides are two quarters of meridians and a quarter of the equator, each of its angles is $90^\circ$ , and its angle sum is $270^\circ$ .

In plain words A spherical triangle is three corners on a ball joined by the straightest paths the ball allows. Its angles are measured right at the corners, along the surface, and they always add up to more than the angles of a flat triangle.

Example Put the center of a sphere of radius $R$ at the origin, with $N = (0, 0, R)$ , $A = (R, 0, 0)$ and $B = (0, R, 0)$ . The side from $N$ to $A$ lies in the plane $y = 0$ , the side from $N$ to $B$ in the plane $x = 0$ and the side from $A$ to $B$ in the plane $z = 0$ . All three planes pass through the center, so all three sides are arcs of great circles, each a quarter of a great circle of length $\pi R/2$ . At $A$ the side toward $N$ leaves in the direction $(0, 0, 1)$ , straight up, and the side toward $B$ leaves in the direction $(0, 1, 0)$ , so they make a right angle. The same happens at $B$ , and at $N$ the two sides leave in the directions $(1, 0, 0)$ and $(0, 1, 0)$ , again at a right angle. The three coordinate planes cut the sphere into eight congruent pieces, and the triangle is one of them, with area $4\pi R^2 / 8 = \pi R^2 / 2$ .

Great circle

Definition A great circle of a sphere is a circle in which the sphere meets a plane through its center. It has the same center and the same radius $R$ as the sphere, so it is the largest circle that can be drawn on the sphere, and it cuts the sphere into two equal hemispheres. Through two points of the sphere that are not opposite ends of a diameter there passes exactly one great circle, the one in the plane through the two points and the center. Through two opposite points there pass infinitely many. Two different great circles always meet, in exactly two opposite points. The equator is a great circle, and so is each full circle through both poles, made of a meridian and the meridian opposite to it. The other circles of latitude are not great circles, because their planes miss the center.

In plain words A great circle is a widest possible belt around a ball, a circle centered at the center of the ball itself. The equator is one and every meridian is half of one, but the other lines of latitude are smaller circles.

Example The circle of latitude $60^\circ$ north lies in the horizontal plane at height $R \sin 60^\circ \approx 0.866R$ above the center. Its center is on the axis at that height, and its radius is $R \cos 60^\circ = R/2$ , half that of a great circle, so its length is $\pi R$ rather than $2\pi R$ . Its plane misses the center of the sphere, so it is not a great circle. Two points of it at longitudes $0^\circ$ and $180^\circ$ lie on exactly one great circle, the circle through both poles in the plane of those two meridians.

Geodesic

Definition A geodesic on a surface is a curve that is as straight as the surface allows. A traveler who walks along it, staying on the surface, never turns to the left or to the right. Every short enough piece of a geodesic is the shortest route on the surface between its ends, although a long geodesic need not be. In a plane the geodesics are the straight lines, and on a sphere they are the great circles and their arcs. Between two points of a sphere that are not opposite, the shortest route on the surface is the shorter arc of the great circle through them. Its length is $R\theta$ , where $\theta$ is the angle in radians between the radii to the two points. The longer arc is also a geodesic but not a shortest route. Between two opposite points every half great circle joining them is a shortest route, of length $\pi R$ . A straight chord through the inside is shorter still, but it leaves the surface, so it does not count. The sides of a spherical triangle are geodesics, which is what makes it a triangle in the geometry of the sphere.

In plain words Geodesics are the straight lines of a curved surface. On a ball they run along great circles, never along the smaller circles of latitude.

Example Take the two points $P$ and $Q$ at latitude $60^\circ$ north and longitudes $0^\circ$ and $180^\circ$ . Walking along their circle of latitude, half of a circle of radius $R/2$ , covers $\pi R/2 \approx 1.571R$ . The great circle through them passes over the North Pole, and each point is $30^\circ$ from the pole, so the shorter arc has angle $60^\circ$ , or $\pi/3$ radians, and length

\frac{\pi R}{3} \approx 1.047R .

The route along the latitude circle is $1.5$ times as long. A traveler on that circle is not going straight, but keeps turning gently toward the pole, which is why the shortest route between two places at the same northern latitude runs north of their circle of latitude.

Spherical excess

Definition The spherical excess of a spherical triangle with angles $\alpha$ , $\beta$ and $\gamma$ , measured in radians, is the amount by which their sum exceeds $\pi$ ,

E = \alpha + \beta + \gamma - \pi .

Girard’s theorem says that on a sphere of radius $R$ the area of the triangle, measured on the curved surface, is

\begin{aligned} A &= R^2 E \\[0.4em] &= R^2 (\alpha + \beta + \gamma - \pi). \end{aligned}

The angles must be in radians and the sides must be great-circle arcs. Since the area is positive and less than half the sphere, $0 < E < 2\pi$ . The theorem is named after Albert Girard, who published it in the seventeenth century. A short proof uses lunes. A lune is the region between two halves of great circles that join the same two opposite points, and a lune whose sides meet at an angle $\alpha$ is the fraction $\alpha/(2\pi)$ of the sphere, with area $2\alpha R^2$ . Extend the sides of the triangle to full great circles. At the vertex with angle $\alpha$ , the two great circles through it form a lune of angle $\alpha$ containing the triangle, and an opposite lune of the same area containing the mirror copy of the triangle through the center. The six lunes from the three vertices have total area $4R^2(\alpha + \beta + \gamma)$ . They cover the triangle and its opposite copy three times each and the rest of the sphere exactly once, so

\begin{aligned} &4R^2(\alpha + \beta + \gamma) \\[0.4em] &\quad = 4\pi R^2 + 4A , \end{aligned}

which gives the formula.

In plain words The extra angle measures area. Every radian of angle beyond $\pi$ stands for an area of $R^2$ , so the bigger a triangle is compared with its sphere, the further its angle sum goes past 180 degrees.

Example For the triangle with three right angles, the angle sum is $3\pi/2$ and the excess is $\pi/2$ , so

A = R^2 \cdot \frac{\pi}{2} = \frac{\pi R^2}{2} ,

which is one eighth of the sphere’s area $4\pi R^2$ , just as the eight octants require. The excess in degrees, $90^\circ$ , cannot be put into the formula, since $90R^2$ would be larger than the whole sphere. With the North Pole and two equator points $45^\circ$ apart instead, the angles are $\pi/2$ , $\pi/2$ and $\pi/4$ , the excess is $\pi/4$ and the area is $\pi R^2/4$ , one sixteenth of the sphere.

Angle sum of a triangle

Definition In a flat plane the three angles of every triangle add up to $180^\circ$ , or $\pi$ radians. The usual proof draws, through one vertex, the line parallel to the opposite side. The two angles this line makes with the other two sides equal the angles of the triangle at the other two vertices, and together with the angle at the chosen vertex they fill a straight angle. The proof needs the parallel postulate of Euclid’s geometry, in the form that through a point not on a line there is exactly one line that never meets it. On a sphere any two great circles meet, so the sphere has no parallel geodesics at all, the proof breaks down, and the angle sum of a spherical triangle with area $A$ on a sphere of radius $R$ is

\alpha + \beta + \gamma = \pi + \frac{A}{R^2} ,

always more than $\pi$ . The sum depends on the size of the triangle compared with the sphere, and it is close to $\pi$ only when the area is small compared with $R^2$ .

In plain words The rule of 180 degrees is a fact about flat surfaces. A small triangle on a big sphere is nearly flat, so its angles add up to nearly 180 degrees, which is why triangles drawn on the ground look perfectly ordinary.

Example Keep the North Pole $N$ and put $A$ and $B$ on two meridians $90^\circ$ apart, each at an angle $\delta$ from the pole, and join $A$ to $B$ by the great-circle arc between them, not by the circle of latitude. The angle at $N$ stays $90^\circ$ , and a calculation with the directions of the sides shows that the angles at $A$ and $B$ are both

\beta = \arccos \frac{\cos \delta}{\sqrt{1 + \cos^2 \delta}} .

$\delta$	angle at $A$ and at $B$	angle sum
$90^\circ$	$90^\circ$	$270^\circ$
$45^\circ$	$54.74^\circ$	$199.47^\circ$
$10^\circ$	$45.44^\circ$	$180.88^\circ$
$1^\circ$	$45.004^\circ$	$180.009^\circ$

As $\delta$ shrinks, the triangle looks more and more like a flat right triangle with two equal sides, whose angles are $90^\circ$ , $45^\circ$ and $45^\circ$ . On the Earth, modeled as a sphere of radius $6371$ kilometers, a triangle with an area of 10,000 square kilometers has an angle sum only about $0.014^\circ$ above $180^\circ$ .

Check yourself

Keep the North Pole as one vertex, but put the two equator vertices $60^\circ$ apart in longitude. Find the three angles, the angle sum, the spherical excess and the area on a sphere of radius $R$ . What fraction of the sphere is the triangle?

Show the answer

Meridians still cross the equator at right angles, so the angles at the two equator vertices are $90^\circ$ , and the angle at the pole equals the difference in longitude, $60^\circ$ . The angle sum is $240^\circ$ . In radians the angles are $\pi/2$ , $\pi/2$ and $\pi/3$ , so the excess is

\frac{\pi}{2} + \frac{\pi}{2} + \frac{\pi}{3} - \pi = \frac{\pi}{3} ,

and the area is $\pi R^2/3$ . That is $1/12$ of the sphere’s area $4\pi R^2$ . As a check, six such triangles fill the northern hemisphere, of area $2\pi R^2$ , and $2\pi R^2/6 = \pi R^2/3$ .

Someone draws a shape from the North Pole down two meridians $90^\circ$ apart to the circle of latitude $45^\circ$ north, and closes it along that circle of latitude. All three corners are right angles, so they use the excess $\pi/2$ and claim the area is $\pi R^2/2$ . What is wrong, and what is the true area of the shape? A cap of height $h$ on a sphere of radius $R$ has area $2\pi R h$ .

Show the answer

The third side runs along a circle of latitude other than the equator, which is not a great circle, so the shape is not a spherical triangle and Girard’s theorem does not apply to it. Its true area comes from the cap above latitude $45^\circ$ north. That circle lies at height $R \sin 45^\circ = R\sqrt{2}/2$ , so the cap has height $R(1 - \sqrt{2}/2)$ and area $2\pi R^2 (1 - \sqrt{2}/2)$ . The two meridians cut off a quarter of the cap, so

\begin{aligned} A &= \frac{\pi R^2}{2}\left(1 - \frac{\sqrt{2}}{2}\right) \\[0.4em] &\approx 0.460 R^2 , \end{aligned}

less than a third of the claimed $\pi R^2/2 \approx 1.571 R^2$ . The genuine spherical triangle on the same three vertices has the great-circle arc as its third side, which bulges north of the circle of latitude. Its angles are $90^\circ$ at the pole and about $54.74^\circ$ at the other two vertices, its excess is about $0.340$ radians, and its area is about $0.340 R^2$ .

Extend the pattern. A regular icosahedron has 20 faces that are equilateral triangles, with five faces meeting at each of its 12 vertices. Place it inside a sphere with all 12 vertices on the sphere, and project each edge from the center of the sphere onto the sphere. Find the angles and the area of each of the 20 spherical triangles you get, and check that the areas add up to the whole sphere. What happens for the regular octahedron, with 8 faces and four of them meeting at each vertex?

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An edge and the center of the sphere lie in one plane through the center, so each projected edge is an arc of a great circle, and the 20 regions are spherical triangles. By symmetry they are congruent, and the five angles around each vertex fill a full turn, so each angle is $360^\circ/5 = 72^\circ$ . The angle sum is $216^\circ$ and the excess is $36^\circ$ , which is $\pi/5$ radians, so each triangle has area $\pi R^2/5$ , and

20 \cdot \frac{\pi R^2}{5} = 4\pi R^2 ,

the area of the whole sphere. For the octahedron four triangles meet at each vertex, so each angle is $90^\circ$ , the excess is $\pi/2$ and each area is $\pi R^2/2$ . These are the eight triangles with three right angles, and $8 \cdot \pi R^2/2 = 4\pi R^2$ again.