likewise from the end of a required sine, or opposite to a required angle, according as a side or an angle is the subject of enquiry *.

(K) Having drawn a perpendicular, agreeably to the foregoing directions; then, if the vertical angle of the oblique triangle be given or sought, find the vertical angle of that rightangled triangle wherein two things are given; but, if the base of the oblique triangle be given or sought, find the base of that right-angled triangle wherein two quantities are given.

Compare the perpendicular, the part given, and the part sought, in that triangle wherein only one quantity is given; find the middle part, and make an equation agreeably to NAPIER'S RULES, marking the term sought with an asterisk (*).

Compare the perpendicular and the similar parts in the triangle where two quantities are given, and make an equation,, which place exactly under the former, and strike out such terms as are common to both the equations.

Turn the remaining quantities into a proportion; thus, if all the terms which are to be struck out of the equations be on the same side of both, put the required term last, that with which it is connected first, and the other two in the middle in any order.

But if the terms which are not struck out of the equations stand on both sides; one unmarked term, in the lower equation, is to the unmarked term exactly above it; as the other unmarked term in the lower equation, is to the marked or required term which stands above it,

(L) CASE I. Suppose two angles at the base of an obliqueangled triangle, and a side opposite to one of these angles were given, to find the vertical angle.

First, find the vertical angle BCD in the triangle BDC, where two entire quantities are given.

B

Here BC is the middle part, and the other parts are the er tremes conjunct. Hence,

*

rad x cos BC cot LBX cot BCD.

*The reason of these rules for drawing a perpendicular is founded on prac tice and observation. For in every right-angled triangle there must be two given quantities, exclusive of the right angle, and it is obvious that the perpendicular must be so drawn as to form one right-angled triangle wherein two quantities are given.

When both the angles at the base are given; if the perpendicular fall without the triangle, it is the most convenient to draw it so that two known parts may come between the perpendicular and the given angle to which it is opposite, hence CD is the proper perpendicular in this case.

Therefore

Therefore cot B: rad: : cos BC: cot BCD.

In the triangle ADC; rad†× cos ▲A=cos DCX sin ACD.
In the triangle BDC; rad X cos BCOS DC X sin BCD.
Therefore cos B: COS LA :: sin BCD: sine ACD.

Had the c been given, and the LA required, we must have proceeded exactly in the same manner, only the A must have been marked instead of ACD, and the proportion would have been thus; sine BCD: sine ACD :: cos B: COS

A.

The sum, or difference, of ACD and BCD, gives ACB according as the perpendicular falls within, or without the triangle.

(M) COROLLARY I. The cosines of the angles at the base, are in proportion to each other as the sines of the angles at the vertex, made by a perpendicular drawn from the vertical angle upon the base.

(N) COROLLARY II. The sum of the cosines of the angles at the base,

Is to their difference;

As the sum of the sines of the angles at the vertex,
Is to their difference. For,

Cos B: cos A :: sine BCD: sine ACD;

And by composition and division,

Cos A+ cos B: COSA COS B :: sine ACD+sine BCD : sine ACD sine BCD.

(0) CASE II. Suppose the two angles at the base of an oblique spherical triangle, and a side opposite to one of these angles were given to find the base.

First find the base DB in the triangle BDC, where two entire quantities are given. Here the complement of CBD is the middle part, and the other parts are the extremes conjunct. Hence,

A

*

d

B

Rad x cos B=tang DBX Cot BC;
Therefore cot BC : rad :: cos B: tang CB,

*

In the triangle ADC, rad × sine AD=cot A × tang DC,
In the triangle, BDC, rad × sine DBcot BX tang DC;

Therefore cot B: cot 4A :: sine DB : sine AD.

Had the base AB been given, and the angle A required, the same method of solution must have been observed, only the

The terms which are to be struck out of the equations are printed in italics.

angle ▲ must have been marked instead of the segment AD, and the last proportion would have been,

Sine DB: sine AD :: cot B: cot A.

The sum, or difference, of AD and DB, gives the base AB, according as the perpendicular falls within, or without the triangle.

(P) COROLLARY I. The co-tangents of the angles at the base are in proportion to each other, as the sines of the segments of the base.

(Q) COROLLARY II. The sum of the co-tangents of the base angles,

Is to their difference;

As the sum of the sines of the segments of the base,
Is to their difference. For,

Cot

cot B: cot LA :: sine DB : sine AD,

And by composition and division,

B+cot A: cot B cot A:: sine DB + sine AD:

sine DB sine AD.

(R) CASE III. Suppose the two angles at the base, and a side opposite to one of them were given, to find the side opposite to the other.

In the triangle ADC, rad x sine DC sine AC X sine ▲A,
In the triangle BDC, rad × sine DC=sine BC X sine B;

Therefore sine 4 A: sine 4 B:: sine BC: sine ac. Had the side AC been given, and the angle A required, the angle A must have been marked, and the proportion would have been, sine AC: sine BC:: sine в : sine A.

(S) COROLLARY I. The sines of the sides, are in proportion to each other as the sines of their opposite angles, et contra. (T) COROLLARY II. The sum of the sines of the sides, Is to their difference;

As the sum of the sines of the angles at the base,

Is to their difference. For,

Sine AC sine BC:: sine 4 B: sine ▲A;

And by composition and division,

4

Sine AC+ sine BC: sine AC 2 sine BC:: sine B+ sine ▲ A: sine LB

sine LA.

(U) CASE IV. Suppose two sides and an angle opposite to one of them were given, to find the base,

First find the segment AD † in the triangle ADC where two entire quantities are given. Here the complement of A is the middle part, and the other two parts are the ex-, tremes conjunct.

*

Hence, rad x cos Acot ACX tang AD;"
Therefore, cot AC: rad :: cós ZA: tang AD.

*

In the triangle BDC, rad x cos BC=cos DC X cos DB,
In the triangle ADC, rad× cos AC=cos DC x cos AD;
Therefore cos AC: COS BC:: cos AD: cos DB.

Had the side AB been given, and BC required, the operation would have been the same, only BC must have been marked instead of DB, then we should have had this proportion:

Cos AD: COS DB:: cos AC : cos BC:

The difference between the base AB and the segment AD gives the segment DB; if AB exceeds AD the perpendicular falls within the triangle, if not, it falls without.

(W) COROLLARY I. The cosines of the sides are in propor tion to each other, as the cosines of the segments, of the base. (X) COROLLARY II. The sum of the cosines of the sides,

Is to their difference;

As the sum of the cosines of the segments of the base,

Is to their difference.

Cos AC: Cos: BC:: cos AD: cos DB;

And by composition and division,

For,

COS AC + cos BC: COS ACCOS BC:: COS AD+cos DB : COS AD

COS DB.

(Y) CASE V. Suppose a side, and its two adjacent angles were given, to find a side opposite to one of these angles.

Find the vertical angle ACD,

in the triangle ADC.

Here the complement of AC is the middle part, and the complements of a and c are the extremes conjunct.

A

D

Hence, rad x cos AC-cot LAX Cot ACD;
"Therefore, cot Arad::cos Ac: cot ACD.

+ When only one angle at the base is given; if the perpendiculat fall without the triangle, let it be drawn so as to fall opposite to that given angle, whether it be acute or obtuse, hence ev is the proper perpendicular in this case.

If

Then the difference between ACB and ACD gives BCD. ACD be less than ACB the perpendicular falls within the triangle, if greater, it falls without.

*

In the triangle BDC, rad x cos BCD tang DCX cot BC,
In the triangle ADC, rad× cos ACD=tang DC X cot ac.
Therefore cOS ACD: COS BCD:: cot AC: cot BC.

Had the side BC been given, and the angle ACB required, the same mode of operation must have been observed, only marking BCD instead of BC, and the proportion would have been thus, Cot Ac: cot BC:: COS ACD: cos BCD.

(Z) COROLLARY I. The cosines of the vertical angles are in proportion to each other, as the co-tangents of the sides. Or the tangents of the sides are reciprocally as the cosines of the vertical angles. For,

Cot AC: cot BC:: cos ACD: COS BCD;

But the tangents are inversely as the co-tangents,

Hence, tang BC: tang AC :: COS ACD: COS BCD.

(A) COROLLARY II. The sum of the co-tangents of the sides, is to their difference; as the sum of the cosines of the vertical angles, is to their difference.

And, The sum of the tangents of the sides,

Is to their difference;

As the sum of the cosines of the vertical angles
Is to their difference. For,

Cot AC: cot BC:: cos ACD: COS BCD,

and tang BC: tang AC:: COS ACD: COS BCD.

By composition and division,

Cot Ac+cot BC: cot Accot BC:: cos ACD+COS BCD : cos ACDCOS BCD, and

Tang BC+tang AC: tang BC 2 tang Ac :: COS ACD+COS BCD :

COS ACD 2 COS BCD.

(B) SCHOLIUM.

From the preceding cases and their corollaries, we deduce the following general rules for solving ten cases of oblique spherical triangles, viz.

I. The sines of the sides are directly proportional to the sines of their opposite angles, et contra.

II. The casines of the vertical angles (made by a perpendicular) are as the co-tangents of their adjacent sides; and the sines thereof, as the cosines of the base angles.

III. The cosines of the segments of the base (made by a perpendicular) are as the cosines of their adjacent sides; and the sines thereof, as the co-tangents of their adjacent base angles.

PROPOSITION