The astronomical triangle

Builds on: Sections 1.1 to 1.4
Reading time: 50 min
Exercises: 6
Olympiad problems: 4

In Section 1.2 we read the position of Sirius off a picture of the sky. From Ankara at nine o'clock on the evening of 15 February, it stood a little east of south, at azimuth 165°, and 32° above the horizon. A telescope on a computer-driven altazimuth mount cannot look at a picture. It knows the declination of Sirius from its catalogue and works out the hour angle from the clock, as in Section 1.3, and before it can turn towards the star it has to convert these two angles into an altitude and an azimuth. The conversion rests on one triangle drawn on the celestial sphere, with its corners at the celestial pole, the zenith and the star, which is called the astronomical triangle. In this section we learn how to solve triangles on a sphere, and then use this one to find where a star stands at any moment, when and where it rises and sets, and how far apart two stars are.

1.5.1Triangles on a sphere

Before you readMake a guess first

On a globe, a triangle has one corner at the North Pole and the other two on the equator, a quarter of the way round the globe from each other. What do its three angles add up to?

A triangle on a sphere has three corners joined by arcs of great circles, the curves that give the shortest route between two points on a sphere (Section 1.1). It is called a spherical triangle. We label the corners A, B and C, and the side opposite each corner with the same letter in lower case, so that side $a$ joins B and C. On the celestial sphere the corners are points such as stars, the pole and the zenith, and the sides are the angular distances between them.

Both the sides and the angles of a spherical triangle are angles. A side is measured by the angle it makes at the centre of the sphere, so side $a$ is the angle BOC, where O is the centre. This is the same as the angular distance from B to C. The angle at a corner is the angle between the two sides that meet there, measured between the directions in which they leave the corner. Every side and every angle is less than 180°.

A small triangle on a sphere is nearly flat, and its angles add up to very nearly 180°. A large one bulges outwards, and its angles add up to more, as the triangle in the guess above shows. Figure 1.13 lets us watch this happen as the triangle grows.

Figure 1.13Triangles on a sphereinteractive

Sides

b

and

c

Angle

A

Sides
Angles
Sum of the angles

The figure shows a sphere with centre O and a triangle ABC drawn on it with heavy arcs of great circles. The thin lines from O to the corners show that each side is measured by the angle it makes at the centre, so that side

a

is the angle BOC. The thin circle completes the great circle through B and C. The sliders set the two sides

b

and

c

, which are kept equal, and the angle A between them, and the table gives the third side, the other two angles and their sum, worked out with the cosine rule below. Tapping gives angles that add up to almost exactly 180°, and gives the 270° of the guess.

A flat triangle obeys the cosine rule of plane geometry, which gives one side from the other two and the angle between them. A spherical triangle has a rule of the same kind, and it is the most useful result in positional astronomy.

ResultCosine rule for a spherical triangle(1.16)

\cos a = \cos b \cos c + \sin b \sin c \cos A

$a$	the side opposite the corner A
$b, c$	the other two sides, which meet at A
$A$	the angle between the sides $b$ and $c$ , at the corner A

The rule holds for each side in turn, for example

\cos b = \cos c \cos a + \sin c \sin a \cos B

. For a very small triangle, with the sides in radians, it becomes the cosine rule of plane geometry,

a^{2} = b^{2} + c^{2} - 2 b c \cos A

Show the derivation

Give the sphere a radius of 1 and put the corner A on the $z$ axis, at the point $(0, 0, 1)$ . Turn the sphere about this axis until B lies in the $x z$ plane. B is at an angle $c$ from A, so it is the point $(\sin c, 0, \cos c)$ . C is at an angle $b$ from A, and the plane OAC makes the angle A with the plane OAB, so C is the point $(\sin b \cos A, \sin b \sin A, \cos b)$ . The side $a$ is the angle between the unit vectors to B and C, and the cosine of the angle between two unit vectors is their scalar product, so

\cos a = \sin c \sin b \cos A + \cos c \cos b,

which is the cosine rule.

Quick checkUse the result above

A spherical triangle has sides $b = 40 °$ and $c = 50 °$ , with an angle of 70° between them. Find the third side $a$ .

degrees

The same rule measures distances on Earth. Two places and the North Pole make a spherical triangle whose sides from the pole run along meridians. Each of these sides is 90° minus the latitude of the place, and the angle between them at the pole is the difference in longitude.

Example 1.12From Hanoi to Ankara

How far is it from Hanoi (21.0° N, 105.8° E) to Ankara (39.9° N, 32.9° E) along the shortest route over Earth's surface? Take Earth's radius to be 6371 km.

Solution

In the triangle formed by the North Pole and the two cities, Hanoi is $90 ° - 21.0 ° = 69.0 °$ from the pole and Ankara is $90 ° - 39.9 ° = 50.1 °$ . The angle at the pole is the difference in longitude, $105.8 ° - 32.9 ° = 72.9 °$ . By (1.16) the side $d$ joining the cities is given by

\begin{aligned} \cos d & = \cos 69.0 ° \cos 50.1 ° \\ + \sin 69.0 ° \sin 50.1 ° \cos 72.9 ° \\ = 0.2299 + 0.2106 = 0.4405, \end{aligned}

so $d = 63.87 °$ . A full circle round Earth is $2 π \times 6371 = 40 030$ km long, so each degree of arc is 111.2 km, and the distance is $63.87 \times 111.2 = 7102$ km.

AnswerThe cities are about 7100 km apart, an arc of 63.9° on Earth's surface.

1.5.2The sine rule and right angles

The cosine rule needs either two sides and the angle between them, or all three sides when we want an angle. A second rule pairs each angle with the side opposite it.

ResultSine rule for a spherical triangle(1.17)

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

$A, B, C$	the angles at the three corners
$a, b, c$	the sides opposite them

An angle found from its sine may be wrong by a supplement, because an angle and its supplement have the same sine, as with

\sin 30 ° = \sin 150 °

. Where an angle could be more than 90°, it is safer to find it with the cosine rule.

Show the derivation

In the coordinates used for the cosine rule, the distance of C from the plane OAB is its $y$ coordinate, $\sin b \sin A$ . The same distance can be found from the corner B instead. C is at an angle $a$ from B, along a great circle whose plane makes the angle B with the plane OBA, so its distance from that plane is $\sin a \sin B$ . Setting the two equal gives $\sin b \sin A = \sin a \sin B$ , which is the first equality of the rule, and the second follows in the same way.

Many problems lead to a triangle with a right angle in it, for example where a route crosses a meridian at right angles, or where a meridian or an hour circle meets the equator. With the right angle at C, the cosine rule for side $c$ and the sine rule become shorter, and together they give a third result that is often needed.

ResultRight-angled spherical triangles(1.18)

\begin{aligned} \cos c & = \cos a \cos b \\ \sin a & = \sin c \sin A \\ \cos A & = \frac{\tan b}{\tan c} \end{aligned}

$c$	the side opposite the right angle at C
$a, b$	the two sides that meet at the right angle
$A$	the angle at the corner A, which lies opposite the side $a$

The first result is the spherical form of Pythagoras' theorem.

Show the derivation

With $C = 90 °$ we have $\cos C = 0$ , and the cosine rule for side $c$ becomes $\cos c = \cos a \cos b$ . With $\sin C = 1$ , the sine rule gives $\sin a = \sin c \sin A$ . For the third result we write the cosine rule for side $a$ ,

\cos a = \cos b \cos c + \sin b \sin c \cos A,

and replace $\cos a$ by $\cos c / \cos b$ from the first result. Solving for $\cos A$ gives

\begin{aligned} \cos A & = \frac{\cos c / \cos b - \cos b \cos c}{\sin b \sin c} \\ = \frac{\cos c (1 - \cos^{2} b)}{\cos b \sin b \sin c} = \frac{\tan b}{\tan c} . \end{aligned}

Quick checkUse the results above

A ship leaves a point on the equator and sails along a great circle, setting off 40° east of north. What is its latitude when it has sailed an arc of 50°?

degrees

1.5.3The astronomical triangle

Before you readMake a guess first

For an observer at latitude 40° N, how far is the zenith from the north celestial pole?

Choose a star X and join it to the north celestial pole P and to the zenith Z by arcs of great circles. With the arc of the meridian from P to Z, these make a spherical triangle PZX, called the astronomical triangle. Each of its sides lies along a circle that we met earlier in this chapter.

The side PZ lies along the meridian. The pole stands at an altitude $φ$ above the north point ((1.11)), so PZ is $90 ° - φ$ .
The side PX lies along the star's hour circle. It is the star's polar distance, $90 ° - δ$ , from (1.7).
The side ZX lies along the star's vertical circle. It is the zenith distance $z = 90 ° - h$ of (1.4).

Two of the angles are also coordinates that we know. At P, the angle between the meridian and the star's hour circle is the hour angle $H$ when the star is west of the meridian. When the star is east of the meridian, the angle is $24^{h} - H$ , which is $- H$ if we write the hour angle as a negative number. At Z, the angle between the meridian, which leaves Z towards the north point, and the star's vertical circle is the azimuth $A$ when the star is east of the meridian, and $360 ° - A$ when it is west. The third angle, at X, is called the parallactic angle, and we will not need it in this chapter.

So one triangle holds the observer's latitude, the star's position in the equatorial system, $δ$ and $H$ , and its position in the horizon system, $h$ and $A$ . If we know the latitude and two of the star's coordinates, the rules of spherical triangles give the other two.

Figure 1.14The astronomical triangleinteractive

Latitude

φ

Hour angle

H

Declination

δ

Watch

Side PZ
Side PX
Side ZX
Angle at P
Angle at Z

The figure shows the celestial sphere seen from outside, from the east-north-east and above the horizon, with the horizon, the meridian and the celestial equator (dotted). The heavy arcs join the north celestial pole P, the zenith Z and a star X into the astronomical triangle, whose sides are

90 ° - φ

90 ° - δ

and the zenith distance

z

. The small arcs mark the angles at P and Z. The sliders set the latitude, the hour angle and the declination, and the table gives every side and angle. When the sky is turned with the button, the angle at P grows while the side PX stays the same, and parts of the triangle on the far side of the sphere are dashed. The figure can be set to show , which Example 1.13 works through, , or , where P is below the horizon and the side PZ is more than 90°.

1.5.4Altitude and azimuth from the hour angle

Before you readMake a guess first

From Ankara, Sirius crosses the meridian at an altitude of 33.4°. How high is it one hour before that?

Suppose we know the observer's latitude $φ$ and a star's declination $δ$ and hour angle $H$ . In the astronomical triangle we then know two sides, $90 ° - φ$ and $90 ° - δ$ , and the angle between them at P. The cosine rule (1.16) gives the third side, the zenith distance:

\begin{aligned} \cos (90 ° - h) \\ = \cos (90 ° - φ) \cos (90 ° - δ) \\ + \sin (90 ° - φ) \sin (90 ° - δ) \cos H . \end{aligned}

For a star east of the meridian the angle at P is $- H$ , but $\cos (- H) = \cos H$ , so the same line holds on both sides. Since $\cos (90 ° - x) = \sin x$ and $\sin (90 ° - x) = \cos x$ , it becomes a formula for the altitude.

ResultAltitude from hour angle and declination(1.19)

\sin h = \sin φ \sin δ + \cos φ \cos δ \cos H

$h$	altitude of the star
$φ$	latitude of the observer, negative in the southern hemisphere
$δ$	declination of the star
$H$	hour angle of the star, turned into degrees at 15° to the hour

On the meridian

H = 0

, and the formula gives

\sin h = \cos (φ - δ)

, which is the altitude at upper culmination of (1.12).

Once the altitude is known, all three sides of the triangle are known, and the cosine rule for the side PX gives the angle at Z. Because $\cos (360 ° - A) = \cos A$ , the azimuth itself can stand in for that angle on either side of the meridian, and the cosine rule becomes $\sin δ = \sin φ \sin h + \cos φ \cos h \cos A$ .

ResultAzimuth from the altitude(1.20)

\cos A = \frac{\sin δ - \sin φ \sin h}{\cos φ \cos h}

$A$	azimuth of the star, measured from north through east

The inverse cosine gives an angle between 0 and 180°. That angle is the azimuth when the star is east of the meridian, with

H

between 12h and 24h. When the star is west of the meridian, with

H

between 0 and 12h, the azimuth is 360° minus that angle.

Example 1.13Sirius over Ankara

At 21:00 on 15 February 2026 the local sidereal time in Ankara ( $φ = 39.9 °$ N) is 5h 54m. The right ascension of Sirius for that year is 6h 46m and its declination is −16.7°. Find the altitude and azimuth of Sirius.

Solution

By (1.10) the hour angle is $H = 5 h 54 m - 6 h 46 m = - 0 h 52 m$ , so Sirius is 52 minutes east of the meridian. In degrees, $H = - 52 \times 15^{'} = - 13.0 °$ . By (1.19),

\begin{aligned} \sin h & = \sin 39.9 ° \sin (- 16.7 °) \\ + \cos 39.9 ° \cos 16.7 ° \cos 13.0 ° \\ = - 0.1843 + 0.7160 = 0.5316, \end{aligned}

so $h = 32.1 °$ . By (1.20),

\begin{aligned} \cos A & = \frac{\sin (- 16.7 °) - \sin 39.9 ° \sin 32.1 °}{\cos 39.9 ° \cos 32.1 °} \\ = \frac{- 0.2874 - 0.3410}{0.6498} = - 0.9671, \end{aligned}

and the inverse cosine gives 165.3°. Sirius is east of the meridian, so this angle is its azimuth.

AnswerSirius is 32.1° above the horizon at azimuth 165.3°, as the panorama in Figure 1.4 shows.

Quick checkUse the result above

From Ankara ( $φ = 39.9 °$ ), how high is Vega ( $δ = + 38.8 °$ ) when its hour angle is 6h?

degrees

1.5.5Rising and setting

Before you readMake a guess first

At sunrise the top edge of the Sun appears, and the air bends its light up over the horizon by about half a degree. By how much do these two effects lengthen the longest day in Ankara?

A star rises or sets when its altitude is zero. Putting $h = 0$ in (1.19) gives the hour angle of setting that Section 1.4 quoted, $\cos H_{0} = - \tan φ \tan δ$ . In practice rising and setting are often counted at an altitude $h_{0}$ a little below the horizon. Refraction in the atmosphere lifts an object near the horizon by about 34′, so a star is seen to rise while its true altitude is still −34′, or −0.57°. Sunrise is counted from the moment the top edge of the Sun's disc appears, and the centre of the disc is then a further 16′ lower, at $- 50^{'} = - 0.83 °$ . Twilight is measured by how far the Sun is below the horizon, which Chapter 2 describes. For any of these, the altitude formula gives the hour angle with $h_{0}$ in place of 0.

ResultHour angle of rising and setting(1.21)

\cos H_{0} = \frac{\sin h_{0} - \sin φ \sin δ}{\cos φ \cos δ}

$H_{0}$	hour angle at which the object sets (it rises at $- H_{0}$ )
$h_{0}$	altitude counted as rising or setting, for example −0.57° for a star seen through the atmosphere and −0.83° for the top edge of the Sun

With

h_{0} = 0

this is (1.15). The Sun's hour angle grows by 15° in each hour of ordinary time, because the Sun returns to the meridian once a solar day, so for the Sun

2 H_{0}

at 15° to the hour is the length of the day in ordinary hours.

Example 1.14The longest day in Ankara

At the June solstice the Sun's declination is +23.44°. How long is the Sun above the horizon in Ankara ( $φ = 39.9 °$ N) on that day? Work it out first for a point-like Sun on an Earth without air, with $h_{0} = 0$ , and then with $h_{0} = - 0.83 °$ .

Solution

With $h_{0} = 0$ , (1.21) gives

\cos H_{0} = - \tan 39.9 ° \tan 23.44 ° = - 0.3625,

so $H_{0} = 111.25 °$ . The Sun is up for $2 H_{0} = 222.51 °$ of hour angle, which at 15° to the hour is 14.83 h, or 14 h 50 min. With $h_{0} = - 0.83 °$ ,

\begin{aligned} \cos H_{0} & = \frac{\sin (- 0.83 °) - \sin 39.9 ° \sin 23.44 °}{\cos 39.9 ° \cos 23.44 °} \\ = \frac{- 0.0145 - 0.2552}{0.7039} = - 0.3831, \end{aligned}

so $H_{0} = 112.53 °$ , and the Sun is up for $2 H_{0} / 15 ° = 15.00$ h.

AnswerWithout air the Sun would be up for 14 h 50 min. Refraction and the size of the Sun's disc add about ten minutes, which makes the longest day in Ankara 15 h 00 min.

The direction in which an object rises follows from (1.20) with $h = 0$ , which leaves $\cos A = \sin δ / \cos φ$ . An object north of the celestial equator rises north of east and sets north of west, and one south of the equator rises and sets south of those points, as Section 1.3 found from the shape of the daily circles.

ResultAzimuth of rising(1.22)

\cos A_{rise} = \frac{\sin δ}{\cos φ}

A_{rise}

azimuth at which the object rises, measured from north through east (it sets at

360 ° - A_{rise}

)

This takes rising to happen at

h = 0

. If the right-hand side is more than 1 or less than −1, the object never rises or never sets.

Example 1.15Sunrise at the June solsticefinish it

In which direction does the Sun rise in Ankara ( $φ = 39.9 °$ N) at the June solstice, when its declination is +23.44°? Give the azimuth.

The start

By (1.22), $\cos A = \sin 23.44 ° / \cos 39.9 °$ .

Your stepNow find the azimuth

A

, and how far north of east the Sun rises.

degrees

1.5.6The distance between two stars

Before you readMake a guess first

Two stars on the celestial equator differ in right ascension by 6h. How far apart are they in the sky?

To find the angular distance between two stars X and Y, we make a triangle with the north celestial pole P. The sides PX and PY are the polar distances $90 ° - δ_{1}$ and $90 ° - δ_{2}$ , and the angle at P between the two hour circles is the difference in right ascension, $α_{1} - α_{2}$ . The cosine rule gives the third side, which is the angular distance $θ$ between the stars.

ResultAngular distance between two stars(1.23)

\cos θ = \sin δ_{1} \sin δ_{2} + \cos δ_{1} \cos δ_{2} \cos (α_{1} - α_{2})

$θ$	angular distance between the two stars
$α_{1}, δ_{1}$	right ascension and declination of the first star
$α_{2}, δ_{2}$	right ascension and declination of the second star, with the difference in right ascension turned into degrees

With latitude in place of declination and longitude in place of right ascension, the same formula gives the distance between two places on Earth, as in Example 1.12.

Example 1.16The Pointers

Merak and Dubhe, the two stars at the end of the bowl of the Plough, point towards Polaris. For the year 2000, Merak has right ascension 11h 01.8m and declination +56.38°, Dubhe has 11h 03.7m and +61.75°, and Polaris has 2h 31.8m and +89.26°. Find the angular distance from Merak to Dubhe and from Dubhe to Polaris.

Solution

From Merak to Dubhe the difference in right ascension is 1.9m, which is $1.9 \times 15^{'} = 0.475 °$ . By (1.23),

\begin{aligned} \cos θ & = \sin 56.38 ° \sin 61.75 ° \\ + \cos 56.38 ° \cos 61.75 ° \cos 0.475 ° \\ = 0.99560, \end{aligned}

so $θ = 5.38 °$ . From Dubhe to Polaris the difference in right ascension is 8h 31.9m, or 127.975°, and

\begin{aligned} \cos θ & = \sin 61.75 ° \sin 89.26 ° \\ + \cos 61.75 ° \cos 89.26 ° \cos 127.975 ° \\ = 0.87706, \end{aligned}

so $θ = 28.7 °$ .

AnswerMerak and Dubhe are 5.4° apart, and Polaris is 28.7° beyond Dubhe, about five times as far. To find Polaris, we follow the line from Merak through Dubhe for about five times the gap between them.

Common errorsSection 1.5

The inverse cosine never gives an angle of more than 180°, so on its own it cannot tell east from west. When the star is west of the meridian, with its hour angle between 0 and 12h, the azimuth is 360° minus the angle that (1.20) gives.
An angle found with the sine rule may really be its supplement, because $\sin x = \sin (180 ° - x)$ . Angles that could be more than 90° should be found with the cosine rule.
Hour angles and differences in right ascension must be turned into degrees before taking a cosine, at 15° to the hour, so a difference of 1h 30m is 22.5°.
For an observer south of the equator $φ$ is negative, the north celestial pole is below the horizon, and the side PZ is more than 90°. The formulas of this section still hold as long as the minus sign is kept.

ExercisesSection 1.56 questions

1.5.1
Which side of the astronomical triangle PZX is equal to $90 ° - φ$ ?
1.5.2
Find the distance along the shortest route over Earth's surface from Ankara (39.9° N, 32.9° E) to Rio de Janeiro (22.9° S, 43.2° W). Take Earth's radius to be 6371 km.
km
1.5.3
From Ankara ( $φ = 39.9 °$ ), find the altitude of Altair ( $δ = + 8.9 °$ ) when its hour angle is 3h.
degrees
1.5.4
At what azimuth does Vega ( $δ = + 38.8 °$ ) rise as seen from Hanoi ( $φ = 21.0 °$ N)?
degrees
1.5.5
At the December solstice the Sun's declination is −23.44°. How long is the Sun above the horizon in Ankara ( $φ = 39.9 °$ N) on that day, counting sunrise and sunset at $h_{0} = - 0.83 °$ ?
hours
1.5.6
A star is seen at azimuth 250°. What can we say about its hour angle?

Olympiad problemsPast IOAA problems, restated4 problems

2013A star over Madrid10 points

Adapted from IOAA 2013 (Volos), theory problem 9. We have restated it in our own words and kept the original data. The solution was last checked on 15 October 2026.

An observer in Madrid, at latitude 40° N, sees a star at a zenith distance of 30° and an azimuth of 50°, with the azimuth measured from the south point. Find the star's hour angle and declination.

I solved it without the solution

AnswerThe star's declination is +18°05′ and its hour angle is 1h 35m, west of the meridian.

2013Vega from two cities10 points

Adapted from IOAA 2013 (Volos), theory problem 14. We have restated it in our own words and kept the original data. The solution was last checked on 15 October 2026.

Vega, at declination +38°47′, is crossing the meridian of Lisbon (longitude 0h 36m W, latitude 39°43′ N) at upper culmination. At that moment, what are its hour angle and its zenith distance as seen from Thessaloniki (longitude 1h 32m E, latitude 40°37′ N)?

I solved it without the solution

AnswerSeen from Thessaloniki, Vega's hour angle is 2h 08m and its zenith distance is 24°33′.

2015From Lima to Yogyakarta

Adapted from IOAA 2015 (Magelang), theory problem 15. We have restated it in our own words and kept the original data. The solution was last checked on 15 October 2026.

An aircraft flies from Lima in Peru (12°02′ S, 77°01′ W) to Yogyakarta in Indonesia (7°47′ S, 110°26′ E) along the shortest route over Earth's surface. Find the latitude of the most southerly point of the route.

I solved it without the solution

AnswerThe route reaches latitude 69°38′ S, far south of both cities.

2018Vega and Altair15 points

Adapted from IOAA 2018 (Beijing), theory problem 10, parts (a) and (b). We have restated it in our own words and kept the original data. The solution was last checked on 15 October 2026.

For the year 2000, Vega has right ascension 18h 36m 56.49s, declination +38°47′07.7″ and parallax 130.23 mas, and Altair has right ascension 19h 50m 47.70s, declination +8°52′13.3″ and parallax 194.95 mas, where 1 mas is a thousandth of an arcsecond.

(a) Find the angular distance between Vega and Altair. [9 points]

(b) Find the distance between the two stars in space, in parsecs. [6 points]

I solved it without the solution

Answer(a) The stars are 34.20° apart in the sky. (b) They are 4.49 pc apart in space, which is closer than either of them is to the Sun.