Case 4, Ambiguous.
One Leg, as

AC = 28 30

= 28 and: an Angle opposite

B=42 34

Given

1. To find the Hypothenuse BC. The Analogy sB : s AC :: R: sBC. That is,

As the Sine of the Angle B=42 3459.8302342
is to the Sine of the Leg AC=28 30=9.6786629
So is the Radius
To the Sine of the Hypoth. BC=44 52=9.8484287

10.

1

10.

2. To find the other Leg B A. The Analogy B : TAC :: 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 so is Radius To the Sine of the Leg BA=36.15=9.7718150

3. To find the other Angle C. The Analogy csCA: R :: csB : SC. That is, As the Co-line of the Leg AC=28 30=9.9438985 is to the Radius

10. so is the Co-fine of the Angle B=42 34=9.8672673 To the Sine of the Angle C=56 57=9.9233688

:

Case 5.

Given The Leg

BA=36 15
AC=28 30

and the Leg

1. To find the Hypothenuse B C.

The Analogy R : CCA :: BA : csBC. That is, As Radius

is to the Co-fine of the Leg ÇA=28 36=9.9438985 To is the Co-line of the Leg BA=36 15-9.9065745 to the Co-line of the Hyp. BG=44 52=9.8504730

2. To find the Angle B.

The Analogy SBA:R:: CA :t A. That is, As the Sine of the Leg BA=36 15=9.7718150 is to the Radius so is the Tangent of the Leg CA=28 30=9.7347644 to the Tangent of the Angle B=42 3459.9629494

10.

3. To find the Anglé C. The Analogy CA: R :: BA : C. That is, As the Sine of the Leg

CA=28 zó=9.6786629 is to Radius fo is the Tangent of the Leg BA=36 1559.8652404 to the Tangent of the Angle C=56 57 ətd.1865775

10.

:

1. To find the Hypothenuse B C. The Analogy 1C : cB :: R:csBC. That is, As the Tangent of the Angle C=58 5=10,1865775 is to the Co-tang, of the Angle B=42 34=10.0370506 fo is the Radius to the Co-line of the Hypoth. BC=44 52=9.8484731

10.

2. To find ibe Leg A B. The Analogy sВ:05C::R: CsBĄ. That is,

8

As the Sine of the Angle B=42, 3439.8302342 is to the Co-sine of the Angle C=56 57=9.7366918 fo is Radius to the Co-line of the Leg BA=36 15=9.9064566

IQ.

3, To find the Leg: A C. The Analogy sC :08B:::R;AC. That is, As the Sine of the Angle C=56 57 59.9233405 is to the Co-line of the Angle B=42 3439.8671673 so is Radius to the Co-sine of the Leg AC=28 30=9.9438268

10.

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

C H. A P.

C H A P. X.

The second Method of Solving Right-angled

Spherical Triangles, by Natural Sines and Tangents.

which belong to the Quantity of the given Side or Angle, in the Column of Sines, or Tangents, as the Analogies shall 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 Thew the Degrees and Minutes contained in the said Arch.

I shall give Examples of the fix absolute Cases, as follows, omitting the Work at large, as unnecessary,

Cafe 1, Given 's The Hypothenuse

and an Oblique Angle, as C = 56 57

BC = 44 52

1. To find the Leg BA. The Analogy R:sC:: SBC:s B A. In Numbers, 100000 : 83819 : : 70545 : 5913©.

But the Number 59130 is the Sine of 38 15zBA.

2. To find the other Leg A C.

The Analogy R :t BC::C8C :1 AC. In Numbers 100000 : 99535 :: 54537 : 54295 : But 54295 is the Natural Tangent 28° 30' = AC.

3. To find the Angle B. The Analogy R:csBC :: 1 C: ct B. In Numbers 100000 : 70875 :: 153692 : 108876. Then 108876 is the Co-tangent of 42° 34'=B.

Case 2.

BC 44 52

Given

The Hypothenuse
and one of the Legs, as BA= 36 15

1. To find the other Leg A C. The Analogy csBA:R::csBC:05 AC. In Numbers 80644 : 100000 :: 70875 : 87881.

Thus 87881 is the Co-sine of 28° 30' = AC.

2. To find the Angle B. The Analogy BC:R::t BA:cs B. In Numbers 99535 : 100000 : : 73323 : 73649 ;

But 73649 is the Co-line of 42° 34' = B.

VOL. II

Q

3. TO