20°

10*

R X cos. AC cos. AB X cos. BC.
R
Cos. AB
47 30' 4"

9.8296742
Cos. BC
32 12

9'9274695

: Cos. AC

55 8

9'7571437

The hypotenųse is acute, because the legs are alike.

2. To find the angle. A.

R X sin. AB = tang. BC X cot. A.
Tang. BC

O'2008431
:R

90
:: Sin. AB
47 30 of

9.8676386

32° 12'

ar. Co.

IO

: Cot. A

40 30 5

10'0684817

The angle A is acute, being of the same affection with the opposite leg.

3. To find the ether angle C.
R X sin. BC = tang. AB X cot. Ç.
Tang. AB

47° 30' 4" 3r. co. 99620356
go

10
:: Sin. BC
32 1 2

9*7266264

:R

: Cot. C

63 58 30

9-6886620

The angle C is acute, being like the opposite leg.
This analogy is like the second.

CASE 6.

EXAMPLE: Given the two angles, namely, A=40* 30' 5'', and C=63° 58' 30'' ; to find the rest.

Projection of the triangle,

1. At the circumference,

Project the angle A, by ste Prob. II. 2. Find y, the pole of the oblique circle ACx; and draw the quadrant ya. Set the angle C from m to n and o ; draw the parallel circle nro ; through r, its intersection with the quadrant, draw the right circle rp, and BCp perpendicular to it.

X

I

2. At the centre.

Project the angle A, bý Prob. II. 1. Set the angle C from * to y and z; and draw the parallel circle yoz; the point of intersection o will be the pole of the oblique circle 1CBq.

B

m

3. In the plane.

dr Project the angle A,

t by Prob. II. 3. Find m, the pole of xACy; draw

X the quadrant mn; set the angle C from r to s and

B t; draw the parallel circle tas ; and through the point z, its intersection with the quadrant mn, draw the oblique circle

f. dzBCf.

Calculation.
1. To find the hypotenuse AC.

R X cos. AC = çot. A X cot. C.
R
: Cot. A

40 30' 5"

10'0684798
:: Cot. C
63 58 30

9'6886626

90°

10*

: Cos. AC

55

8

97571424 +

The

* The log. of the 4th term, in this and the third analogy, gives more than half a second beside 55° 8', on account of the fraction of a second, neglected in other parts of the triangle.

The hypotenuse is acute, the angles being of the same kind.

RX cos. C =sin. A X cos. AB.
Sin. A 40° 30' 5" ar. co. Oʻ1874433
go

10*
:: Cos. C 63 58 30

996422303

:R

: Cos. AB
47 30 4

908296738 The leg AB is acute, like the opposite angle:

3. To find the other leg BC.
R X cos. A = sin. C X coś. BC.
Sin. C 63° 58' 30' ar. co. Oʻ0464323

90
:: Cos. A.
40 30 5

9:8810365

:R

The leg BC is acute, like the opposite angle.
This analogy is like the second.

RECTILATERAL SPHERICAL

TRIGONOMETRY.

If the supplemental triangle to a rectangular spherical triangle be drawn, by. Theor. IV. General Properties of Spherical Triangles, one side will be a quadrant, being the supplement of a right angle ; and therefore it will be a

Quadrant

quadrantal or réctilateral spherical triangle. And as the sine, cosine and tangent of an arc are the same as the sine, cosine and tangent of its supplement respectively, the equations for the rectangular are all equally applicable to its supplemental

B tectilateral triangle. Hence a rectilateral spherical triangle ABC may be solved like a rectangular one by the method of five circular parts. The quadrantal side, called the quadrant, or radius, like the right angle, is not considered as separating the adjacent parts. The two angles adjacent to the quadrant, the complement of the other angle, opposite to the quadrant and called the hypotenusal angle, and the complements of the other two sides are to be regarded as the five circular parts.

Thus, in the triangle ABC, AB is the quadrant, or radius, the angle A, the complement of AC, the complement of the hypotenusal angle C, the complement of CB and the angle B are the circular parts.

The ambiguous cases are the same as in rectangular triangles, that is, when one side and its opposite angle are given.

OBLIQUE SPHERICAL TRIG

ONOMETRY.

THEOREM I.

In a spherical triangle, if the angles at the base be of the same affection, the perpendicular, falling on the base