F from an angle of a spherical triangle there be drawn a perpendicular to the oppofite fide, or base, the rectangle contained by the tangents of half the fum, and of half the difference of the fegments of the bafe is equal to the rectangle contained by the tangents of half the fum, and of half the difference of the two fides of the triangle.

Let ABC be a spherical triangle, and let the arch CD be drawn from the angle C at right angles to the base AB, tan 1⁄2 (BD + AD) x tan 1⁄2 (BD—AD)=tan 1⁄2 (BC+AC) × tan

(BC AC).

Because (26.), cof BC: cof AC:: cof BD: cof AD, (E. 5.) cof BC+cof AC: cof BC-cof AC:: cof BD+cof AD: cof BD- cof AD. But, (1. Cor. 3. Pl. Tr.),

cof BC + cof AC: cof BC cof AC:: cot (BC + AC): tan (BC-AC); and also, cof BD + cof AD: cof BD — cof AD: : cot (BD+AD): tan 1⁄2 (BD—AD), therefore (11. 5.) cot (BC+AC): tan (BC—AC) : : cot 1⁄2 (BD + AD) : tan (BD-AD). And because rectangles of the fame altitude are as their bafes, therefore, tan (BC+AC) ×' cot (BC + AC) : tan (BC + AC) × tan 1⁄2 (BC — AC) : : tan 1⁄2 (BD + AD) × cot (BD+AD); tan1⁄2 (BD + AD) × tan (BD-AD). But tan (BC+AC)xcot (BC+AC) R2, (1. Cor. def. 9. Pl. Tr.) and also, tan1⁄2 (BD+AD)×

cot

tan

1

1

1

(BD+AD)=R2, therefore (9. 5.) tan (BD+AD)x (BD-AD) = tan (BC + AC) × tan (BC—AC), 1⁄2 1⁄2

Q. E. D.

A a 3

COR.

COR. 1. Because the fides of equal rectangles are reciprocally proportional, tan1⁄2 (BD+AD) : tan 1⁄2 (BC + AC) : : tan (BC — AC): tan (BD — AD).

COR. 2. Since, when the perpendicular CD falls within the triangle, BD + AD AB, the bafe; and when CD falls without the triangle BD-ADAB, therefore in the firit cafe, the proportion in the laft corollary becomes, tan(AB): tan (BC+AC) :: tan (BC-AC): tan (BD-AD); and in the fecond cafe, it becomes by inverfion and alternation, tan (AB): tan (BC + AC) : : tan 1⁄2 (BC—AC); tan (BD + AD).

THE preceding propofition, which is very useful in fpheri cal trigonometry, may be eafily remembered from its analogy to the propofition in plane trigonometry, that the rectangle under half the fum, and half the difference of the fides of a plane triangle, is equal to the rectangle under half the fum, and half the difference of the fegments of the bafe. See (G. 6.), alfo 3d Cafe Pl. Tr. We are indebted to NAPIER for this and the two following theorems, which are fo well adapted

adapted to calculation by Logarithms, that they must be confidered as three of the most valuable propofitions in Trigonometry.

IF

PRO P. XXX.

F a perpendicular be drawn from an angle of a spherical triangle to the oppofite fide or bafe, the fine of the fum of the angles at the base is to the fine of their difference as the tangent of half the base to the tangent of half the difference of its fegments, when the perpendicular falls within; but as the cotangent of half the bafe to the co-tangent of half the fum of the fegments, when the perpendicular falls without the triangle: And the fine of the fum of the two fides is to the fine of their difference as the cotangent of half the angle contained by the fides, to the tangent of half the difference of the angles which the perpendicular makes with the fame fides, when it falls within, or to the tangent of half the fum of these angles, when it falls without the triangle.

If ABC be a spherical triangle, and AD a perpendicular to the base BC, fin (C+B) : fin (C-B) : : tan

tan

BC :

(BD-DC), when AD falls within the triangle; but

AD

B

B

D

BG : cot 1⁄2 (BD+DC),

when

when AD falls without. And again, fin (AB+ AC) : fin (AB-AC) :: tan BAC : tan (BAD-CAD), when AD falls within; but when AD falls without the triangle, fin (AB+AC): fin (AB-AC) :: cot BAC :

For in the triangle BAC (27.), tan B: tan C:: fin CD: fin BD, and therefore, (E. 5.), tan C + tan B: tan C-tan B :: fin BD + fin CD: fin BD- fin CD. Now, (dividing the firft fides of the laft two equations in page 324. by cof A x cof B, and Sect. III. 8. Pl. Tr.) tan C +tan B: tan C-tan B:: fin (C+B): fin (C—B), and fin BD + fin CD fin BD- fin CD: tan tan (BD-CD), (3. Pl. Trig.), therefore, because ratios which are equal to the fame ratio are equal to one another (11. 5.), fin (C+B): fin (C-B) :: tan1⁄2 (BD+CD) :

tan1⁄2

(BD+CD):

But when AD is within the triangle, BD+CD BC, and therefore fin (C+B): fin (C—B) : : tan1⁄2 BC: tan1⁄2 (BD—CD). And again, when AD is without the triangle, BD-CD = BC, and therefore fin (C+B): fin (CB): tan1⁄2 (BD+CD): tan BC, or because the tangents of any two arches are reciprocally as their co-tangents, fin (C+B): fin (C-B): : cot BG: cot (BD+CD),

The second part of the propofition is next to be demonstrated. Because (28.) tan AB; tan AC: cofCAD: cof BAD, tạn AB +

:

tan AB+ tan AC: tan AB-tan AC :: cof CAD+ cof BAD : cof CAD- cof BAD. But tan AB + tan AC: tan AB tan AC: fin (AB+AC) : fin (AB-AC), and (1. Cor. 3. Pl. Trig.), cof CAD + cof BAD : cof CAD- cof BAD: : cot (BAD+CAD): tan (BAD-CAD). Therefore (11. 5.) fin (AB+AC) : fin (AB—AC) : :

cot1⁄2 (BAD+CAD) : tan 1⁄2 (BAD-CAD). Now, when AD is within the triangle, BAD+CAD — BAC, and therefore fin (AB+AC) : fin (AB—AC) : : cot — BAC ;

But, if AD be without the triangle, BAD-CAD BAC, and therefore fin (AB+AC) : fin (AB—AC) : :

cot

1

(BAD+CAD): tan BAC; or because cot (BAD + CAD): tan BAC : : cot BAC : tan

1

(BAD+CAD), fin (AB+AC) : fin (AB—AC); ; cot BAC: tan (BAD+CAD). Wherefore, &c. E. D:

PROP.