A'P' C, it is sufficient to solve the small right angled spherical triangle ABC.

To the half lune AP'B, we add the triangle ABC, and we have the quadrantal triangle AP' C; and by subtracting the same from the equal half lune APB, we have the quadrantal triangle PAC.

When we have the side, AC, of the same triangle, we have its supplement, A' C, which is a side of the triangle A'PC, and of A'P'C. When we have the side, CB, of the small triangle, by adding it to 90°, we have P'C, a side of the triangle A'P'C; and subtracting it from 90°, we have PC, a side of the triangle APC, and A'PC.

EXAMPLES.

1. In a quadrantal triangle, there are given the quadrantal side, 90°, a side adjacent, 42° 21′, and the angle opposite this last side, equal to 36° 31′. Required the other parts.

By this enumeration we cannot decide whether the triangle APC or AP' C, is the one required, for AC=42° 21′ belongs equally to both triangles. The angle APC AP' C=36° 31'=AB.

We operate wholly on the triangle ABC.

To find the side BC, call it the middle part.

R sin.BC tan.AB cot.A CB.

We now have all the sides, and all the angles of the four triangles in question.

2. In a quadrantal spherical triangle, having given the quadrantal side, 90°, an adjacent side, 115°, 09', and the included angle, 115° 55', to find the other parts.

This enunciation clearly points out the particular triangle A'P'C. A'P'=90°; and conceive A'C=115° 09'. Then the angle P'A'C-115° 55'=P'D.

From the angle P'A' C take 90° or P'A'B, and the remainder is the angle OA'D=BAC =25° 55'.

We here again operate on the triangle ABC. A'C, taken from 180°, gives

To find BC, we call it the middle part.

R sin.BC sin. AC sin. BAC.

To find AB we call it the middle part.

R sin.AB=tan. BC cot. BAC.

To find the angle C, we call it the middle part.

R cos. C cot.ĄC tan.BC

Thus we have found the side P' C=113° 18′ 19′′

The angle A'P' C=117° 33′ 52"
66 P' CA'=101° 40′ 7′′

Ans.

3. In a quadrantal triangle, given the quadrantal side, 90°, a side adjacent, 67° 3', and the included angle, 49° 18', to find the other parts.

Ans. The remaining side is 53° 5′ 46′′, the angle opposite the quadrantal side, 108° 32′ 27′′, and the remaining angle, 60° 48′ 54′′.

4. In a quadrantal triangle, given the quadrantal side, 90°, one angle adjacent, 118° 40′ 36′′, and the side opposite this last mentioned angle, 113° 2′ 28", to find the other parts.

Ans. The remaining side is 54° 38′ 57′′, the angle opposite, 51° 2′ 35′′, and the angle opposite the quadrantal side is 72° 26′ 21′′.

5. In a quadrantal triangle, given the quadrantal side, 90, and the two adjacent angles, one 69° 13′ 46", the other 72° 12′ 4′′, to find the other parts.

Ans. One of the remaining sides is 70° 8′ 39′′, the other is 73° 17′ 29", and the angle opposite the quadrantal side is 96° 13′ 23′′.

6. In a quadrantal triangle, given the quadrantal side, 90°, one adjacent side, 86° 14′ 40′′, and the angle opposite to that side, 37° 12′ 20′′, to find the other parts.

Ans. The remaining side is 4° 43′ 2′′, the angle opposite, 2° 51′ 23′′, and the angle opposite the quadrantal side, 142° 42′ 2′′.

7. In a quadrantal triangle, given the quadrantal side, 90°, and the other two sides, one 118° 32′ 16′′, the other 67° 48′ 40′′, to find the other parts-the three angles.

Ans. The angles are 64° 32′ 21′′, 121° 3′ 40′′, and 77° 11′ 6′′; greater angle opposite the greater side, of course.

the

8. In a quadrantal triangle, given the quadrantal side, 90°, the angle opposite, 104° 41′ 17′′, and one adjacent side, 73° 21′ 6′′, to find the other parts.

Ans. The remaining side is 49° 42′ 18′′, and the remaining angles are 47° 32′ 39′′, and 67° 56′ 13′′.

OBLIQUE ANGLED SPHERICAL

TRIGONOMETRY.

ALL cases of oblique angled spherical trigonometry may be solved by right angled trigonometry, except two; because every oblique angled spherical triangle is composed of the sum or difference of two right angled spherical triangles.

When a side and two of the angles, or an angle and two of the sides are given, to find the other parts, conform to the following directions : Let a perpendicular be drawn from an extremity of a given side, and opposite a given angle or its supplement; this will form two right angled spherical triangles; and one of them will have its hypotenuse and one of its adjacent angles given, from which all its other parts can be computed; and some of these parts will become as known parts to the other triangle, from which all its parts can be computed.

To facilitate these computations, we here give a summary of the practical truths demonstrated in the foregoing propositions.

1. The sines of the sides of spherical triangles are proportional to the sines of their opposite angles.

2. The sines of the segments of the base, made by a perpendicular from the opposite angle, are proportional to the cotangents of their adjacent angles.

3. The cosines of the segments of the base are proportional to the cosines of the adjacent sides of the triangle.

4. The tangents of the segments of the base are proportional to the the tangents of the opposite segments of the vertical angles.

5. The cosines of the angles at the base, are proportional to the sines of the corresponding segments of the vertical angles.

6. The cosines of the segments of the vertical angles are proportional to the cotangents of the adjoining sides of the triangle.

The two cases in which right angled triangles are not used, are,
1st. When the three sides are given to find the angles; and,
2d. When the three angles are given to find the sides.

The first of these cases is the most important of all, and for that reason great attention has been given to it, and two series of equations, (T) and (U), have been deduced to facilitate its solution.

We now apply the following equation to find the angle A, of the triangle ABC, whose sides are a, b, c. a=70° 4′ 18′′. b=63° 21′ 27′′. c=59° 16′ 23′′. a is opposite A, b is opposite B. and c is opposite C.

We write the second member of this equation thus:

R

10; and

is the sin.c subtracted from 10, which we call

sin.c

When we apply the equation to find the angle A, we write a first, at the top of the column; when we apply the equation to find the angle B, we write b at the top of the column. Thus,