The Analogy SB s AC: R: SBC. That is,

As the Sine of the Angle.

is to the Sine of the Leg So is the Radius

B=42 34 9.8302342 AC=28 30=9.6786629

10.

To the Sine of the Hypoth. BC=44 52=9.8484287

2. To find the other Leg B A.

The Analogy tBtAC:: R: SBA. That is,

As the Tangent of the Angle B=42 34=9.9629494 is to the Tangent of the Leg AC=28 30=9.7347644 fo is Radius

To the Sine of the Leg

་

IO.

BA=36.15=9.7718150

3. To find the other Angle C.

The Analogy cs CA: R: csB SC. That is,

୪

As the Co-fine of the Leg AC 28 30=9.9438985 is to the Radius fo is the Co-fine of the Angle B=42 34=9.8672673

To the Sine of the Angle

.IO.

C=56 57=9.9233688

BA=36 15
AC 28 30

1. To find the Hypothenufe BC.

The Analogy R : csCA :: ésBA : csBC. That is, As Radius

IO.

is to the Co-fine of the Leg CA=28 30=9.9438985 fo is the Co-fine of the Leg BA=36 159.9065745 to the Co-fine of the Hyp. BC 44 52=9.8504730

2. To find the Angle B.

The Analogy sBA: R: tCA: tA. That is,

As the Sine of the Leg is to the Radius

BA=36 15=9.7718150

10.

fo is the Tangent of the Leg CA=28 30=9.7347644 to the Tangent of the Angle B=42 34=9.9629494

3. To find the Angle C.

The Analogy CA: R:: tBA tC. That is,

As the Sine of the Leg

is to Radius

CA=28 36=9.6786 6 29

10.

fo is the Tangent of the Leg BA=36 15=9.8652404 to the Tangent of the Angle C 56 57=10.1865775

1. To find the Hypothenufe B C.

The Analogy tC ctB:: R: csBC. That is,

:

8

As the Tangent of the Angle C-56 57-10,1865775 is to the Co-tang, of the Angle B=42 34 10.0370506 fo is the Radius

10.

to the Co-fine of the Hypoth. BC=44 52=9.8484731

2. To find the Leg AB.

The Analogy s B: csC:: R: csBA. That is,

As the Sine of the Angle

B=42, 349.8302342 is to the Co-fine of the Angle C-56 57-9.7866918 fo is Radius

IQ.

to the Co-fine of the Leg BA=36 15=9.9064566

3. To find the Leg: AC.

The Analogy sC : csB:: R; cs AC. That is, As the Sine of the Angle C=56 C=56 57-9.9233405 is to the Co-fine of the Angle B42 34 9.8671673 fo is Radius

10.

to the Co-fine of the Leg AC-28 30=9.9438268

Thus I have fhewn how by the Logarithms all the Varieties of Cafes in a Right-angled Spherical Triangle are folved; and that in fo natural an Order, that none can be puzzled with any Difficulty in this Method, which, of all others, is far the best.

CH.A P.

CHA P. X.

The fecond Method of Solving Right-angled Spherical Triangles, by Natural Sines and Tangents.

I

N the Canon or Tables of Natural Sines and Tangents, you are only to take out those Numbers which belong to the Quantity of the given Side or Angle, in the Column of Sines, or Tangents, as the Analogies fhall direct, and by the Golden Rule, you find the Natural Sine or Tangent of the Arch required; which being fought in the Table, will there fhew the Degrees and Minutes contained in the faid Arch.

I fhall give Examples of the fix abfolute Cafes, as follows, omitting the Work at large, as unneceffary.

Cafe 1.

The Hypothenufe

Given

an

8 :

BC= 44 52

and an Oblique Angle, as C = 56 57

1. To find the Leg B A.

The Analogy R: SC::s BC: SBA.

In Numbers, 100000 83819: 70545 59130.

But the Number 59130 is the Sine of 36 15-BA.

2. To find the other Leg A C.

The Analogy RtBC::cs Ct AC. In Numbers 100000 : 99535:: 54537,54295: But 54295 is the Natural Tangent 28° 30′ AC.

3. To find the Angle B.

The Analogy Rcs BCC

ct B.

In Numbers 100000: 70875 :: 153692: 108876. Then 108876 is the Co-tangent of 42° 34' B.

Cafe 2.

Given

The Hypothenufe

BC 44 52

and one of the Legs, as BA = 36 15

1. To find the other Leg A C.

The Analogy csBA: R::cs BC: cs AC.

In Numbers 80644: 100000 :: 70875
Thus 87881 is the Co-fine of 28° 30′

2. To find the Angle B.

87881. AC.

The Analogy tBC: RtBA: cs B.

In Numbers 99535: 100000 :: 73323
But 73649 is the Co-fine of 42° 34'

73649;

B.

3. To