If we substitute, in equation (b), for A and B, the values

} (a+b) and 1c, and also substitute s for (a+b+c) and sc for (a+b-c), equation (b) will become

cos 1 (a+b) — cos 1⁄2 c

tan stan (s- c).

(d)

cos 1 (a+b)+cos 1 c

(e)

Comparing (a), (c), and (d), we obtain

cot (2 C-E) tan E=tans tan 1 (s — c).

By beginning with the second equation of [49], and treating

it in the same way, we obtain as the result,

(f)

tan 4 (2 C — E) tan E=tan (s — a) tan 1 (s —b). † By taking the product of (e) and (f), we obtain the elegant formula,

tan2E=tan stan (s-a) tan (s-b) tan (s-c), [52] which is known as l'Huilier's Formula.

By means of it E may be computed from the three sides, and then the area of the triangle may be found by [51].

III. In all other cases, the area may be found by first solving the triangle so far as to obtain the angles or the sides, whichever may be more convenient, and then applying [51] or [52].

If, therefore, we know the radius of the sphere, we can express the area of a spherical triangle in the ordinary units of area.

* See Wentworth & Hill's Tables, page 20,

1. Given A= 84° 20′ 19′′, B = 27° 22′ 40′′, C=75° 33′ ; find E=26159", F=0.12682 R2.

2. Given a 69° 15′ 6′′, b=120° 42' 47", c=159° 18′ 33′′; find E=216° 40′ 28′′.

3. Given a = = 33° 1′ 45′′, b = 155° 5′ 18′′, C= 110° 10′; find E=133° 48′ 53′′.

4. Find the area of a triangle on the earth's surface (regarded as spherical), if each side of the triangle is equal to 1°. (Radius of earth = 3958 miles.)

CHAPTER IX.

APPLICATIONS OF SPHERICAL TRIGONOMETRY.

§ 63. PROBLEM.

To reduce an angle measured in space to the horizon.
Let O (Fig. 48) be the position of the observer on the ground,

H

C

B

N

FIG. 48.

B'

A

R

AOB=h, the angle measured in space, (for example, the angle between the tops of two church spires), OA' and OB' the projections of the sides of the angle upon the horizontal plane HR, AOA' m and BOB': =n, the angles of inclination of OA and OB respectively to the horizon. Required the angle A'OB' = x made by the projections on the horizon.

The planes of, the angles of inclination AOA' and BOB' produced intersect in the line OC, which is perpendicular to the horizontal plane (Geom. § 520).

From O as a centre describe a sphere, and let its surface cut the edges of the trihedral angle O-ABC in the points M, N, and P. In the spherical triangle MNP the three sides. MN=h, MP : 90° — m, NP 90° — n, are known, and the spherical angle P is equal to the required angle x.

From 48 we obtain

cos x = √cos s cos (sh) sec m secn,

where (m+n+h)=s.

§ 64. PROBLEM.

To find the distance between two places on the earth's surface (regarded as spherical), given the latitudes of the places and the difference of their longitudes.

P

Let M and N (Fig. 49) be the places; then their distance MN is an arc of the great circle passing through the places. Let P be the pole, AB the equator. The arcs MR and NS are the latitudes of the places, and the

arc RS, or the angle MPN, is the difference of their longitudes. Let MR = a, NS=b, RS=1; then in the spherical triangle MNP two sides, MP

=90°-a, NP-90°-b, and the

FIG. 49.

M

B

R

included angle MPN= l, are given, and we have (from § 56) tan m cot a cos l,

cos MN= sin a sec m sin (b+m).

From these equations first find m, then the arc MN, and then reduce MN to geographical miles, of which there are 60 in each degree.

§ 65. THE CELESTIAL SPHERE.

The Celestial Sphere is an imaginary sphere of indefinite radius, upon the concave surface of which all the heavenly bodies appear to be situated.

The Celestial Equator, or Equinoctial, is the great circle in which the plane of the earth's equator produced intersects the surface of the celestial sphere.

The Poles of the equinoctial are the points where the earth's axis produced cuts the surface of the celestial sphere.

The Celestial Meridian of an observer is the great circle in which the plane of his terrestrial meridian produced meets the surface of the celestial sphere.

Hour Circles, or Circles of Declination, are great circles. passing through the poles, and perpendicular to the equinoctial.

The Horizon of an observer is the great circle in which a plane tangent to the earth's surface, at the place where he is, meets the surface of the celestial sphere.

The Zenith of an observer is that pole of his horizon which is exactly above his head.

Vertical Circles are great circles passing through the zenith of an observer, and perpendicular to his horizon.

The vertical circle passing through the east and west points of the horizon is called the Prime Vertical; that passing through the north and south points coincides with the celestial meridian.

The Ecliptic is a great circle of the celestial sphere, apparently traversed by the sun in one year from west to east, in consequence of the motion of the earth around the sun.

The Equinoxes are the points where the ecliptic cuts the equinoctial. They are distinguished as the Vernal equinox and the Autumnal equinox; the sun in his annual journey passes through the former on March 21, and through the latter on September 21.

Circles of Latitude are great circles passing through the poles of the ecliptic, and perpendicular to the plane of the ecliptic.

The angle which the ecliptic makes with the equinoctial is called the obliquity of the ecliptic; it is equal to 23° 27', nearly, and is often denoted by the letter e.

These definitions are illustrated in Figs. 50 and 51. In Fig. 50, AVBU is the equinoctial, P and P' its poles, NPZS the celestial meridian of an observer, NESW his horizon, Z his zenith, M a star, PMP' the hour circle passing through the star, ZMDZ the vertical through the star.