trigonometry, 
823 
Or, As radius is to the cotangent of either 
i of the acute angles, so is the side opposite to 
that angle, to the side adjacent to it ; and vice 
: versa. 
It may also be observed, that the sine of either 
of the acute angles of a right-angled triangle, 
being equal to the cosine of the other, the lat- 
ter may be used instead of the former, when- 
ever it renders the operation more simple. 
Example I. In any right-angled plane triangle 
ABC, 
C BC 321 ? Required the other 
‘ Glven 1 A B 53° 7' 4S" 5 parts. 
By Construction. Make RC — 324, and 
\ B = .73° 1 ' ; then raise the perpendicular CA, 
meeting BA in A ; and the triangle is construct- 
I ed : in which AB will be found to measure 540, 
I and AC 432 ; and A A, which is the comple- 
j ment of A B, is 36° 53'. 
Br Calculation. 
* Rad. or sine - 90° - 10.0000000 
* Tan. A B - 55° T 48" 10. 1 249371 
;; Side BC - 324 - 2.5105450 
l Side AC - 432 - 2.6354821 
* Sine A A or cos. A B 53° 7' 48" 9 . 7781524 
: Side BC * - - 324 - 2.5105450 
** Rad. or sine AC - 90° - 10.0000000 
l Side AB 540 - 2.732392 6 
And 90° — 53° 7' 48" = 36° 52' 12" A A. 
We shall now give in a tabular form, (1) The 
solution of the cases of right-angled plane tri- 
angles : and (2) The solution of the cases of ob- 
lique plane triangles. 
( The Solution of the Cases of right-angled plane Tri- 
angles, fig. 248. 
Case] 
Given. 
Sought. 
Proportion. 
1 
The hy- 
pothenu. 
AC and 
theangles 
The leg 
BC. 
A.sthe radius (or the sine 
of B) is to the hyp. AC ; 
so is the sine of A, to its 
opposite side BC. 
2 
The hy- 
poth. AC 
and one 
leg AB. 
The 
angles. 
As, AC ; rad. * * AB ( 
sine of C ; whose comple- 
ment gives the angle A. 
3 
The hy- 
po t h AC 
and one 
leg AB. 
The 
other leg 
BC. 
Let the angles be found 
by case 2 ; then, as rad. * 
AC * * sine of A ; BC. 
4 
The an- 
gles and 
one leg 
AB. 
The hy- 
pothe- 
nuse AC. 
As, sine of C ( AB * * 
rad. (sine of B) [ AC. 
5 
The an- 
gles and 
one leg 
AB. 
The 
other leg 
BC. 
As, sine of C * AB * * 
sine of A * BC. 
Or, rad. * tang, of A * * 
AB : BC. 
€ 
The two 
legs AB 
and BC. 
The 
angles. 
As, AB ‘ BC * ‘ rad. • 
tang, of A, whose com- 
plement gives the angle 
C. 
7 
The two 
legs AB 
and BC. 
The hy- 
pothe- 
nuse AC. 
Find the angles, by -case 
6, and from thence the 
hyp. AC, by case 4 . 
The Solution of the Cases of Oblique Plane Triangles, [ 
fig. 249, 
| Case! 
Given. 
Sought. 
Proportion. 
1 
The an- 
gles and 
one side 
AB. 
Either of 
the other 
sides, sup- 
pose BC. 
As, sine of C * AB * * 
sine of A * BC. 
2 
Twosides 
AB, BC, 
and the 
ang.C op- 
posite to 
1 of them. 
The other 
angles A 
and ABC. 
As, AB * sine of C * * 
BC l sine of A ; which, 
added to C, and the sum 
subtracted from i£Q (gives 
the angle ABC. 
3 
Twosides 
AB, BC, 
and the 
ang.C op- 
posite to 
I of them. 
Theother 
side AC. 
Find the angle ABC, 
by case 2 ; then, as sine 
of A * BC [ ‘ sine of ABC 
: AC. 
4 
Twosides 
AB, AC ; 
and the 
included 
angle A. 
Theother 
angles C, 
and ABC. 
As, A3 -f AC .* AB — 
AC * * tang. of the comp.of 
■|A * tang, of an ang. which 
added to the said comp, 
gives the greater ang. C ; 
and subtracted leaves the 
lesser ABC. 
5 
Twosides 
AB, AC, 
and the 
included 
angle A. 
Theother 
side BC. 
Find the angles, by case 
4, and then BC, by ease 1. 
6 
All the 
sides. 
An an- 
gle, sup- 
pose A. 
Let fall a perpendi- 
cular BD, opposite the 
required angle, and sup- 
pose DG — AD ; then 
ac : bc-j-ba :: bc — 
BA ; CG, which subtract- 
ed from AC, and the re- 
mainder divided by 2, 
gives AD ; whence A will 
be found, by case 2 of 
right angles. 
Spherical Trigonometry. Spherical Tri- 
gonometry is the art whereby, from three 'given 
parts of a spherical triangle, we discover the 
rest : and, like plane trigonometry, is either 
right-angled or oblique-angled. But before we 
give the analogies for the solution of the several 
cases in eit er, it will be proper to premise the 
following theorems : 
Theorem I, In all right-angled spherical tri- 
angles, the sign of the hypothenuse \ radius ” 
sine of a leg ’ sine of its opposite angle. And 
the sine of a leg ; radius ; * tangent of the other 
leg * tangent of its opposite angle. 
Demon 4 ration. Let EDAFG (fig. 250,) repre- 
sent the eighth part of a sphere, where the quad- 
rantal plaqes EDFG, EDBC, are both perpen- 
dicular to the quadrantal plane ADFB .- ana the 
quadrantal plane ADGC is perpendicular to the 
plane EDFG; and the spherical triangle ABC 
is right-angled at B, where CA is the hypothe- 
nuse, and BA, BC, are the legs. 
To the arches GF, CB,draw the tangents HF, 
OB, and the sines GM, Cl, on the radii DF, DB; 
also draw BL the sine of the arch AB, and CK 
the sine of AC : and then join IK and OL. Now 
HF, GB, GM, Cl, are all perpendicular to the 
plane ADFB. And HD, GK, OL, lie all in the 
same plane ADGC. Also FD, IK, BL, lie all in 
the same plane ADGC. Therefore the right- 
angled triangles HFD, CIK, ODL, having the 
equal angles HDF, CKI, OLB; are similar. And 
CK * DG ** Cl * GM ; that is, as the sine cf 
the hypotheaute * rad. \ \ sine of a leg ] «ne of 
its opposite angle. For GM is the sine of the 
arc OF, which measures the angle CAB. Also, 
LB * DF ** BO * FH: that is, as the sine of a 
leg ' radius ” tangent of the other leg ; tan- 
gent of its opposite angle, q. e. d. 
Hence it follows, that the sines of the angles 
of any oblique spherical triangle ACD (fig. 251.) 
are to one another, directly, as the sines of the 
opposite sides. Hence it also follows, that in right- 
angled spherical triangles, having the same per- 
pendicular, the sines of the bases will be to each 
other, inversely, as the tangents of the angle* at 
the bases. 
Theorem II. In any right-anp-led spherical 
triangle ABC (fig. 252.) it will be, As radius ij 
to the co-sine of one leg, so ••• the co sine of the 
other leg to the co-sine of the hypothenuse. 
Hence, if two right-angled spherical triangb-, 
ABC, CBD, ( fig. 251,) have the same perpen- 
dicular BC, the co-sines of their hvpothenu-.-s 
will be to each other, directly, as the co-sines of 
their bases. 
Iheoiem III. In any spherical triangle it 
will be, As radius is to the sine cf either an-.-h* 
so is the co-sine of the adjacent leg to the co- 
sine of the opposite angle. 
Hence, in right-angled spherical triangles, 
having the same perpendicular, the co-sines of 
the angles at the base will be to each other, di- 
rectly, as the sines of the ver :cal angles. 
. Theorem IV. In any right-angled spherical 
triangle it will he, As radius is to the co-sine of 
the hypothenuse, so is the tangent of either an- 
gle to the co-tangent of the other angle. 
As the sum of the sines of two unequal arches 
is to their difference, so is the tangent of half 
the sum of those arches to the tangent of half 
their difference : and as the sum of the co-sides 
is to their difference, so is the co-tangent of half 
the sum of the arches to the tangent of half the 
difference of the same arches. 
Theorem V. In any spherical triangle ABC 
(figs. 253 and 254), it will be, As the co-tangent 
of half the sum to half their difference, so ^ 
the co-tangent of half the bast; to the tangent 
of the distance (DF) of the perpendicular from 
the middle of the bas.e. 
Since the last proportion, by permutation 
, AC -J- BC 
becomes co-tang. • co-tang. AE v 
AC — BC 
DF, and as the tangents 
of any two arches are, inversely, as their co- 
tangents ; it follows, therefore, that tang AE * 
. AC -j- BC AC — BC 
tang. -- . , tang. - — — ’ tang. DF • 
or, that the tangent of half the base is to the 
tangent of half the sum of the sides, as the tan- 
gem of half the difference of the sides to the 
tangent of the distance of the perpendicular 
from the middle of the base. 
Theorem VI. In any spherical triangle ABC 
(fig. 253), it will be, As the co-tangent of half 
the sum of the angles at the base is to the tan- 
gent of half their difference, so is the tangent 
of half the vertical angle to the tangent of the 
angle which the perpendicular CD makes with 
the line CF bisecting the vertical angle. 
The following propositions and remarks, con- 
cerning spherical triangles (selected and com- 
municated to Dr. Hutton by the reverend Ne- 
vil Maskelyne, D. D. astrondmer-royal, F. R, s.) 
will also render the calculation of them perpicu- 
ous, and free from ambiguity. 
1. A spherical triangle is equilateral, iscscelar, 
or scalene, according as i; has its three angles’ 
all equal, or two of them equal, or all three un- 
equal ; and vice versa. 
2. The greatest side is always opposite the 
