822 
TRIGONOMETRY. 
find a side, begin with an angle opposite t® a 
given side. 
Note. When two sides and an angle opposite 
to one of them are given, to find the rest, the 
question is sometimes ambiguous, or admits of 
two different answers. 
Thus, if the given angle is opposite to the 
least of the two given sides, the angle to be 
found, by the rule; may be either an acute an- 
gle or its supplement; but, if it is opposite to 
the greater side, the required angle will be 
acute. 
Example I. In the plane triangle ABC, fig. 247. 
C EC 23 6? yards. Required^the 
Given -? BC 350 3 other parts. 
CAB 38° 40' 
By Construction. 
1. Lay down the line BC = 350, from some 
convenient scale of equal parts. 
2 . Make the Z B =: 38° 40' by a scale of 
chords, or other instrument. 
3. With the centre C, and radius 236, taken 
from the same scale of equal parts, cross BA in 
A or a. 
4. Join CA or C a, and the triangle ABC, or 
«BC, is the one required. 
Then, the angles C and A, measured by the 
scale of chords, and the side BA, or B a, by the 
scale of equal parts, wiil be found to be as fol- 
lows, viz. 
Z C 291° I Z A 67|° I AB 184 
or 73^ | or 112 j° J or 362 
By Calculation. 
As side AC - 236 2.3729120 
Is to sine Z B 
So is side BC 
38° 40' 
350 
To sine Z A 67° 54' or 112 ° 6 ' 
38° 40' 38° 40' 
7.6270880 
9.7957330 
2.5440680 
9-9668890 
Sum 106° 34' or 150° 46' 
Subtract 180° O' ISO 0 0' 
Leaves 
Then, 
* Sine Z B 38° 40 
73° 26' or 29° 14' Z C 
• 9.7957330 
Side AC 236 
Sine Z C 29° 14' 
Side AB 184.47 
0.2042670 
- 2.3729120 
- 9.6887467 
- 2.2659257 
Or, 
Sine ZB 38° 40' - 9.7957330 
Side AC 236 
Sine Z C 73° 26' 
Side aB 3S2.04 
0.2042670 
- 2.3729120 
- 9.9815S70 
- 2.5587660 
As the results in this rule are determined by' 
means of the sines, which are always the same 
for an acute angle and its supplement, it is plain 
that, in certain cases, there may be two trian- 
gles with the same data ; one acute-angled, and 
the other obtuse-angled : and, consequently, 
when there is no restriction or limitation in the 
question, either of them may be taken for the 
tine required. 
Thus, in the figure given above, where the 
least side AC is opposite to the given acute an- 
gle B, it appears, from the construction, that 
either ABC or cBC is the triangle sought. But, 
when the given angle is right or obtuse, it will 
be opposite to the greatest side, and in this case 
there can be no ambiguity : for then neither of 
the other angles can be obtuse, and the geome- 
trical construction will accordingly form only 
one triangle. 
Instrum ent ally. 
In the first proportion, extend the compasses 
from 236 to 350, upon the line of numbers, and 
that extent will reach, upon the sines, from 
38|° to 67f°, for the Z. A. 
In the second proportion, extend from 38|.° 
to 29Y= or V3|° upon the sines, and that extent 
will reach, upon the line of numbers, from 236 
to 184, or 362, for the side AB, or aB. 
Example II. In the plane triangle ABC, 
C AB 131 f AC 88.045 
Given X BC 97 Ans. |zA 47° 46' 
tZC 90° CZ B 42° 14' 
Required the other parts. 
Example III. In the plane triangle ABC, 
T BC 305 f AC 237.93 
Given X ZB 51° 15' Ans. 2 AB 185.09 
L Z C 37° 21; CZA91°24' 
Required the other parts. 
Example IV. In the plane triangle ABC, 
r AB 195 f z. A 92° 13' 
Given X AC 203 Ans. 4 Z C 42° 47' 
CZ B 45° C BC 286.87 
Required the other parts. 
Example V. In the plane triangle ABC, 
C BC 345 f AB 174.07 
Given X AC 232 j or 374.56 
CZB 37° 20' Ans. j Z C 27° 4' 
1 or 78° 16' 
| Z A 115° 36' 
Required the other parts. or 64° 24^ 
Case II. When two sides and their included 
angle are given, to find the rest. 
Rule. As the sum of any two sides of a plane 
triangle, is to their difference, so is the tan rent 
of half the sum of their opposite angles, to the 
tangent of half thjsir difference. 
Then the half difference of these angles, added 
to their half sum, gives the greater angle, and 
subtracted from it gives the less. 
And as all the angles are now known, the 
remaining side may be found by Case I. 
Note. Instead of the tangent of half the sum 
of the two unknown angles, we may use the co- 
tangent of half the given angle, or the tangent 
of half its supplement, which are all equal to 
each other. 
Example. In any plane triangle ABC, 
r AB 1075? , „ „ . . , 
Given X BC 2391$ teet ' Re fi U!r ed the rest. 
CZB 34° 46' 
By Construction. 1 . Draw BC = 2394, 
from a scale of equal parts. 
2. Set off the Z B = 34° 46', by a scale of 
chords, or other instrument. 
3. Make AB = 1075, by the same scale of 
equal parts, as before. 
4. Join A, C, and the triangle Is constructed. 
Then, the parts being measured, we shall have 
Z A= 123"-g-, ZC = 22 tV, and side AC = 
1630 feet. 
By Calculation. 
AB -f EC 3469 
AB c/s BC 1319 
m A -4” C . , 
Tan. — - — 72° 37' 
„ A CO C 
1 an. 50° 32' 
3 . 5402043 
6.4597957 
3.1202448 
10.5043702 
10.0844107 
Sum 123° 9 ' Z. A 
Diff. 22° 5' ZC. 
Then , 
I Sine ZA 123° 9' or 56° 51' 9.9228509 
Side BC - • 2394 
'Sine ZB - 34° 46' 
Side AC - 1(630.5 
0.0771491 
3.3791241 
9.7560544 
3.2123276 
Case III. When the three sides arc given, to 
find the angles. 
Rule. Make the longest side the base, and let 
fall a perpendicular upon it from the opposite 
angle. 
Then, as the base, or sum of its segments,, is 
to the sum of the other two sides, so is the dif- 
ference of those sides, to the difference of the 
segments of the base. 
And half this difference, being added to half 
the base, will give the greater segment ; and, 
subtracted from it, will give the less. ' 
Then, in each of the right-angled triangles, 
formed by the perpendicular, there wilf be* 
known two sides and an angle opposite to one 
of them; from whence the other angles may be 
found, by Case I. 
Example I. In any plane triangle ABC, 
C AB 464 ? , _ . , , 
Given -< AC 348 > y ar<R- Required the 
C BC 690 
angles. 
By Construction. 1 . Draw BC= 690, by 
a scale of equal parts. ’ J 
2. With the centres B, C, and radii 464 and 
348, taken from the same scale, describe arcs 
intersecting each other in A. 
3. Join AB, AC, and the triangle is con- 
structed. 
Then, by measuring the angles with a pro- 
tractor, or by the scale of chords, they will be 
found to be nearly as follows, viz. ZA — 1 1 A G 1 
ZB = 27°, and Z.C = 37°|. 
By Calculation. 
Having let fall the perpendicular AD, it will be 
» BC or BD —[—DC 690 - 2.8388491 
AB -f- AC 
AB co AC 
BD co DC 
812 - 
116 - 
7 . 161 1509 
2.90 95560 
2.0644580 
136.51 - 2.1351649 
Hence 
«90 -f 136.51 
= 413.25 — BD. 
. , 690 — 136.51 
And =: 276 . 75 = CD. 
Then, in the triangle ABD, right-angled at D, 
AB 
BD 
Sine ZD - 
Sine ZBAD 
464 - 2.6665180 
413.25 - 2.6162129 
90° - 10.0000000 
62° 57' - 
90° O' 
9.94969-19 
27° 3' Z. B. 
And, in the triangle ACD, right-angled at D, 
; AC - 348 - 2.5415792 
’ DC - 276.75 - .2.4420876 
” Sine ZD - 90° 0 ' - 10.0000000 
Sine Z CAD 52° 40' 
90° O' 
9 . 9005084 
Also 
And 
37° 20 ' Z C 
62° 57' Z. BAD 
52° 40' Z. CAD 
Makes 1 15° 37' Z. BAC. 
Whence Z B — 27° 3', ZC = 37° 25' and 
Z BAC — 1 15° 37'. 
These three problems include all the cases or 
varieties of plane triangles, as well right-angled 
as oblique, that can possibly happen ; but there 
are some other theorems, for right-angled tri- 
angles, that are often more convenient in prac- 
tice than the general ones, the most useful of 
which is the one that follows : 
Case IV. In any right-angled triangle, As 
radius is to. the tangent of either of the acute 
angles, so is the side adjacent to that angle, to 
the side opposite to it ; and vice versa. 
