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In a right angled spherical triangle given the hypothenuse 64° 40', and an adjacent angle = 64° 35'; to find the rest.

=

Example IV.

Given one leg 42° 12′, and its opposite angle = 48°; to find the rest.

Given a leg to find the rest.

Example V.

54° 48′, and its adjacent angle = 48°;

Example VI.

Given the two legs 54° 19′; and 42o 12′, respectively; to find the rest.

Example VII,

Given the two oblique angles = 48° and 64° 35′ respectively; to find the rest.

Example VIII.

Given a quadrantal side, one of the other sides = 115° 9', and the angle comprehended between them 115° 55'; to find the rest.

Ans. Angles 101° 4′ and 117° 34', side 113° 18'.

Example IX.

Given in an oblique angled spherical triangle, the

side a 44° 13′ 45′′, b = 84° 14′ 29′′, and their included angle c = 36° 45′ 28′′; to find the rest. This example corresponds with case 2, A prob. 2, of oblique angled spherical triangles; and may first be solved by means of the subsidiary are, in the manner there explained.

Thus, first find, so that

tan cos c tan b.

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D

cos 36° 45′ 28′′.... 9.9037261
tan 84° 14′ 29′′... 10.9963395

tan tan 82° 49′ 33′′....10.9000656

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This arco exceeds a, therefore the perpendicular AD from the vertical angle falls on the base produced: hence the 2d expression becomes

COS C

cos b cos (a)

cos b

Hence, to log cos 84° 14′ 29′′

9.0014632

COB 38° 35′ 48′′ .... 9.8929604

add

from the sum

18.8944236

take cos 82° 49′ 33′′.... 9-0965132

Rem. cos ccos 51° 6' 11".... 9-7979104

To find the remaining parts use the known proportion of the sines of sides to the sines of their opposite angles; thus

As sin c

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51° 6' 11" .... 9·8911340

36° 45′ 28′′

44° 13′ 45′′

....9.7770158

....9.8435629
....9-7294447

32° 26′ 7′′....

And so is sin b.. 84° 14' 29" 9.9978028

....

To sin B.......130° 5′ 21′′. .9.8836846

Here the logarithmic sine 9.8836846 answers either to 49° 54′ 39′′ or to its supplement 130° 5′ 21′′; the

former of which is the exterior angle ABD, the latter the angle B of the triangle.

2d Method, by Napier's Analogies.

Taking the 14th and 15th formulæ at the end of sect. 4, of the preceding chapter, we have

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The log. computation will therefore stand thus:
To log cot c....... 18° 22′ 44′′..........
... 10-4785395
Add log sin ( — a) 20° 0′ 22′′.... 9.5341789

20-0127184

From the sum
Take log sin (b + a) 64° 14′ 7′′.... 9.9545255

Rem.logtan (BA) 48° 49′ 38′′.... 10.0581929

Also, to log cot c.. 18° 22' 44"....10-4785395 Add log cos (b — a) 20° 0′ 22′′.... 9.9729690

20.4515085

From the sum
Take log cos (b + a) 64° 14′ 7′′.... 96381663

Rem.logtan (B+ A) 81° 15′ 44′′

....10.8133422

Hence 81° 15′ 44′′ + 48° 49′ 38" 130° 5′ 22′′ = B, and 81° 15′ 44′′ 48° 49′ 38′′ = 32°26′ 6′′ = A; agreeing nearly with the result of the former computation.

Then to find c, use the proportion, as sin A: sin a :: sin c sin csin 51°6′12′′.

Here it would seem, from a comparison of the methods, that the first is rather quickest in operation, while the last is probably the easiest to remember, and provides best against the occasions of ambiguity.

Example X.

In an oblique angled spherical triangle ABC, given the side c = 114° 30', side a = 56° 40′, and the angle c opposite the first side 125° 20': to find the rest. Ans. A 48° 30′, в = 62° 54′, b = 83° 12'.

Example XI,

Given A = 48° 30′, c = 125° 20′, c = 114° 30′; to find the rest.

Example XII.

Given a = 56° 40′, c = 114° 30′, в = 62° 54; to find the rest.

Given A find the rest.

Example XIII.

48° 30′, c = 125° 20′, b = 83° 12′; to

Example XIV.

Given a 56° 40, b = 83° 12′, c = 114° 30'; to find the rest.

Example XV.

Given A 48° 30′, в = 62° 54′, c = 125° 20′; to find the rest.

**For more examples see chap. x.

CHAPTER VIII.

On Projections of the Sphere,

SECTION I.

Astronomical Definitions.

1. SINCE the figure of the earth differs but little from that of a sphere, it is usual in the greater part of the

inquiries and computations of astronomers, to proceed as though it were a sphere in reality; and since, to an observer on the earth, the heavens appear as a very large concave sphere, every part of which is equidistant from him, it has been found expedient to imagine various lines and circles to be described upon the earth, and the planes of several of them to be extended every way until they mark other similar lines and circles upon the imaginary concave sphere of the heavens. Some of these it now becomes necessary to explain.

2. The axis of the earth is an imaginary right line passing through the centre, about which line it is supposed to turn uniformly once in a natural day.

3. The extremities of this axis are called the poles of the earth.

4. That great circle of the earth, the poles of which are the poles of the earth, is called the equator.

5. If the axis of the earth be supposed produced both ways to the concave heavens, it is then called the axis of the heavens; its extremities are called the poles of the heavens; and the circumference formed by extending the plane of the equator to the celestial concavity is called the celestial equator, or the equinoctial.

6. A secondary to the equator drawn through any place on the earth, and passing through the poles, is called the meridian of that place.

7. The latitude of any place upon the surface of the earth, is its distance from the equator measured on an arc of the meridian passing through it. A less circle passing through any place parallel to the equator is called a parallel of latitude. Places that lie between the equator and the north pole have north latitude; if they lie between the equator and the south pole they have south latitude.

8. All places that lie under the same meridian have the same longitude; and those places which lie under different meridians have different longitudes. The dif ference of longitude between any two places, is the dis

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