(5) Given the right ascension and declination of a star; to find its latitude and longitude.

Let V be the vernal equinox, S the star, VD, VL the the equator and the ecliptic, SD, SL perpendicular to VD, VL.

Then

= α, SD = dec

From (1), is determined; and from (2) and (3), λ and 7 are determined.

Ex. Given the right ascension of a star 5h 6m 42o.01, and its declination 45° 51' 20".1 N.; to find its longitude and latitude, the obliquity of the ecliptic being 23° 27′ 19′′.45.

EXAMPLES.

1. Find the apparent time of sunrise at a place whose latitude is 40° 42', when the sun's declination is 17° 49' N. Ans. 4h 56m.

2. Given the latitude of a place = 40° 36′ 23′′.9, the hour angle of a star=46° 40′4′′.5, and its declination=23° 4′ 24′′.3; to find its azimuth and altitude.

Ans. Azimuth = 80° 23′ 4′′.47, altitude = 47° 15'′ 18′′.3.

3. Find the altitude and azimuth of a star to an observer in latitude 38° 53′ N., when the hour angle of the star is 3h 15m 20s W., and the declination is 12° 42' N.

Ans. Altitude =

39° 38' 0"; azimuth = S. 72° 28' 14" W.

4. Given the latitudes of New York City and Liverpool 40° 42' 44" N. and 53° 25' N., respectively, and their longitudes 74° 0' 24" W. and 3° W., respectively; to find the shortest distance on the earth's surface between them in miles, considering the earth as a perfect sphere whose radius is 3956 miles.

NOTE.This is evidently a case of (4) where two sides and the included angle are given, to find the third side. Ans. 3305 miles.

5. The latitudes of Paris and Pekin are 48° 50' 14" N. and 39° 54' 13" N., and their difference of longitude is 114° 7' 30"; find the distance between them in degrees.

Ans. 73° 56' 40".

GEODESY.

226. The Chordal Triangle. Given two sides and the included angle of a spherical triangle; to find the corresponding angle of the chordal triangle.

The chordal triangle is the triangle formed by the chords of the sides of a spherical triangle.

Let ABC be a spherical triangle, O the centre of the sphere, A'BC the colunar triangle, and M, N the middle points of the arcs A'B, A'C. Then the chord AB is M parallel to the radius OM, since they are both perpendicular to the chord A'B. Similarly, AC is parallel to ON.

In the spherical triangle A'MN, we have cos MN

N

Α

=cos A'N cos A'M+sin A'N sin A'M cos A' (Art. 191) Denote the angle BAC of the chordal triangle by A1. Then arc MN or angle MONA, A'N = 1 (π — b), A'M = {(π — c), and A'= A.

.. cos A1 = sin b sinc+ cos b cos c cos A. (1)

1

with similar values for cos B1 and cos C1.

Cor. 1. If the sides b and c are small compared with the radius of the sphere, A, will not differ much from A.

cos A1 = cos A+ sin A, nearly.

But sin 16 sin c = sin2 4 (b + c) — sin2 1 (b -- c),

and cos bcos c = cos2 1 (b + c) — sin2 1 (b − c).

Substituting in (1) and reducing, we get

=

tan A sin2 (b+c)-cot A sin2 (b-c). (2) which is the circular measure of the excess of an angle of the spherical triangle over the corresponding angle of the chordal triangle.

The value in seconds is obtained by dividing the circular measure by the circular measure of one second, or, approximately, by the sine of one second.

Cor. 2. The angles of the chordal triangle are, respectively, equal to the arcs joining the middle points of the sides of the colunar triangles.

227. Legendre's Theorem. If the sides of a spherical triangle be small compared with the radius of the sphere, then each angle of the spherical triangle exceeds by one-third of the spherical excess the corresponding angle of the plane triangle, the sides of which are of the same length as the arcs of the spherical triangle.

Let a, b, c be the lengths of the sides of the spherical triangle, and the radius of the sphere; then the circular a b c measures of the sides are respectively-, -,

*

Hence,

=

̧ b2 + c2 — a2 _ 2 b2c2 +2c2a2+2a2b2 — ab*c*..

—

(1)

γ

*The term centre of the sphere; and similarly for

is the circular measure of the angle which the arc a subtends at the

b с
and

r

Now if A', B', C' denote the angles of the plane triangle whose sides are a, b, c, respectively, we have

Let A A'+0, where 0 is a very small quantity; then

=

where ▲ denotes the area of the plane triangle whose sides

are a, b, c.

or

A+B+C = == spherical excess (Art. 219)

... A—A'=B-B'=C-C'=

=

32

spherical excess.

Cor. 1. If the sides of a spherical triangle be very small compared with the radius of the sphere, the area of the spherical triangle is approximately equal to