Had the side AB been given, and BC required, the operation would have been the same, only вc 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.

(K) COROLLARY I. The cosines of the sides are in proportion to each other, as the cosines of the segments, of the base. (L) COROLLARY II. The sum of the cosines of the sides, Is to their difference;

As the sum of the cosines of the seg

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

And by composition and division,

Cos AC+COS BC: COS AC~ COS BC::COS AD + COS DB : COS AD

~COS DB.

(M) 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.

Hence, rad x cos AC cot LAX Cot ACD;

Therefore, cot Arad::cos ac : cot ACD.

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 DC X 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.

(N) 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;

N

But the tangents are inversely as the co-tangents,

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

(0) 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 BED.
By composition and division,

Cot AC+ cot BC: cot AC Cot BC::cos ACD+COS BCD: cos ACD~COS BCD, and

Tang BC+tang AC: tang BC~tang AC::COS ACD + COS BCD÷

COS ACDCOS BCD.

(P) 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 cosines 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 XXVI.

(Q) In any spherical triangle. I. If the perpendicular fall within the triangle,

The co-tangent of half the sum of the two sides,

E D A EB

Is to tangent of half their difference;
As the co-tangent of half the base,
Is to the tangent of the distance of a perpendicular from the middle
of the base. II. Or, If the perpendicular fall without the triangle,

Tangent of half the difference of the sides, is to co-tangent of half their sum; as tangent of half the base, is to co-tangent of the distance of a perpendicular from the middle of the base. DEMONSTRATION. Cos AC + cos BC COS AC~ COS BC::COS AD+COS DB COS AD~COS DB (L. 177.) But,

Cos AC+ COS BC cos ACCOS BC:: co-tang (AC+BC) tang

(ACBC) (R. 112.) And,

Cos AD + COS BD: COS AD~COS BD:: co-tang (AD+DB): tang (ADDB) (R..112.)

Co-tang (AC+BC): tang (AC~BC):: co-tang ✈ (AD+ DB) tang (ADDB.)

But (AD+DB)=AE, and (AD~ DB) DE, hence co-tang (AC+BC): tang (AC~BC):: co-tang AE tang DE.

But when the perpendicular falls without the triangle, 1⁄2 (AD~DB)= AE and (AD+DB)=de.

Co-tang (AC + BC): tang (AC~ BC):: co-tang DE: tang AE; and by inversion, tang (ACBC): co-tang (AC+BC) tang AE co-tang DE. Q. E. D.

(R) COROLLARY I. The distance of a perpendicular from the middle of the base, or as some writers call it the altern, or alternate base, is always equal to half the difference of the segments of the base, when the perpendicular falls within the triangle; or equal to half the sum of the segments, when a perpendicular falls without the triangle. Either of the above rules will bring the same conclusion, whether the perpendicular falls within or without the triangle; only in the first case, the fourth number will be less than half the base, and in the second case, it will be greater.

That the rules are both the same may be shewn thus,

When the perpendicular falls within the triangle.

Co-tang (AC+BC) tang (AC~BC):: co-tang AE tang DE; and inversely, tang (ACBC): co-tang (AC+BC):: tang DE co-tang AE,

But the tangents are reciprocally as their co-tangents.

... Tang (ACBC): co-tang (AC + BC) :: tang AE Cotang DE; the same conclusion as when the perpendicular falls without the triangle.

(S) COROLLARY II. The tangent of half the base,

Is to the tangent of half the sum of the sides;

As the tangent of half the difference of the sides,

Is to the tangent of the distance of a perpendicular from the middle of the base. Or,

The tangent of half the sum of the sides, is to the co-tangent of half their difference; as the tangent of half the base, is to the co-tangent of the distance of a perpendicular from the middle of the base. And,

According as this distance is less or greater than half the base, the perpendicular falls within, or without the triangle. For we have already shewn,

Co-tangent (AC+BC): tang (AC~BC):: co-tang AE : tang DE; and tang (AC~BC): co-tang (AC+BC):: tang AE cotang DE.

But, since the tangents are reciprocally as the co-tangents, Tang AE tang (AC+BC)::tang (AC~BC): tang DE; and tang (AC+BC): co-tang (AC~BC)::tang AE co-tang

DE.

(T) COROLLARY III. If the triangle be isosceles, or equilateral, the perpendicular will fall on the middle of the base, except the sides be quadrants, and then it may fall in any part of the base.

(U) COROLLARY IV. In a right-angled triangle, the rectangle of the tangents of half the sum and half the difference, of the hypothenuse and one leg, is equal to the square of the tangent of half the other leg. For in this case B and D will coincide, and DE will be equal to AB, or equal AE.

PROPOSITION XXVII. (See the Fig. to PROP. XXVI.)
(W) In any spherical triangle,

I. If the perpendicular fall within the triangle.
The co-tangent of half the sum of the angles at the base,
Is to the tangent of half their difference;

As the tangent of half the vertical angle,

Is to the tangent of the excess of the greater of the two vertical angles, made by a perpendicular, above half the aforesaid vertical angle.

II. Or, If the perpendicular fall without the triangle.

The tangent of half the difference between the base angles,
Is to co-tangent of half their sum;

As the co-tangent of half the vertical angle,

Is to the co-tangent of the excess of the greater of the two vertical angles, formed by a perpendicular upon the base, above half the aforesaid vertical angle.

DEMONSTRATION. COS A+ Cos B: COS A COS B:: Sine ACD + sine BCD sine ACD~sine BCD. (B. 175.)

·Cos A+cos B: cos A~cos B: cot (A + B): tang 1 (A~B) (R. 112.) Sine ACD + sine BCD : sine ACD~. sine BCD::tang (ACD +BCD) tang (ACD~BCD) (P. 111.) Therefore,

Cot (A+B) tang (A~B):: tang (ACD+BCD): tang (ACD~ BCD.)

But (ACD+BCD)=ACB, and (ACD~BCD)=ECD, hence cot (A+B) tang (A~B):: tang ACB tang ECD.

But when the perpendicular falls without the triangle, (ACD-BCD) ACB, and ECD (ACD+BCD.)

Cot (AB) tang (A + B) :: tang (ACD - BCD): tang (ACD+BCD); by inversion, tang (A+B): cot (A ~B) ;; tang ECD tang ACB.

But the tangents and co-tangents are reciprocally proportional.

Tang (AB): cot (A+B):: cot ACB cot. ECD. Q.E.D.

(X) COROLLARY I. The excess of the greater of the two vertical angles (formed by a perpendicular) above half the vertical angle, is equal to half the difference of those vertical angles, when the perpendicular falls within the triangle; or half their sum, when it falls without. Either of the above rules will bring the same conclusion, whether the perpendicular falls within or without the triangle; only in the former case, the fourth number will be less than half the vertical angle, and in the latter it will be greater. That the rules are the same may be shewn thus, When the perpendicular falls without the triangle. Tang (AB) co-tang (A + B)::cot ACB cot ECD; by inversion, cot (A + B): tang (A~B)::cot ECD: cot

ACB.

But the tangents are reciprocally as the co-tangents.. ·.· Cot (A+B): tang 1⁄2 (A~B): tang ACB tang ECD, the same expression as when the perpendicular falls within the triangle.

(Y) COROLLARY II. The co-tangent of half the sum of the angles at the base, is to the tangent of half their difference; as the tangent of half the vertical angle, is to tangent of half the difference between the two vertical angles, formed by a perpendicular, or to tangent of half their sum, according as the perpendicular falls within or without the triangle.

CHAP. V.

INVESTIGATION OF GENERAL RULES FOR CALCULATING THE SIDES AND ANGLES OF OBLIQUE-ANGLED SPHERICAL TRIANGLES WITHOUT MAKING USE OF A PERPENDICULAR.

PROPOSITION XXVIII.*

(Z) If the cosine of any side of a spherical triangle be multiplied by the radius, and the rectangle of the cosines of the other two sides be deducted from the product; the remainder divided by the rectangle of the sines of these two sides, will be equal to the cosine of the included angle divided by the radius,

* Legendre's Geometry, 6th Edition, page 386, et seq.