PROBLEM II.

In any oblique angled spherical triangle, of the three sides and three angles, any three being given, it is required to find the other three.

In this Table, the references (c. 4.), (c. 5.), &c. are to the cases in the preceding Table, (16.), (27.), &c. to the propositions in spherical Trigonometry.

other angles

SOLUTION.

Let fall the perpendicular CD
from the unknown angle, not
required, on AB.

One of the R: cos A:: tan AC: tan AD, (c.
2.); therefore BD is known,
and sin BD : sin AD: : tan A:
tan B, (27.); B and A are of
the same or different affection,
according as AB is greater or
less than BD, (16.)

B.

The third

side

BC.

Let fall the perpendicular CD
from one of the unknown an-
gles on the side AB.
Rcos A:: tan AC: tan AD, (c.
2.); therefore BD is known,
and cos AD: cos BD: : cos AC:
cos BC, (26); according as the
segments AD and DB are of
the same or different affection,
AC and CB will be of the same
or different affection.

Two sides

AC and BC,

The angle Sin BC: sin AC:: sin A: sin B,

B

opposite to
the other
given side
AC.

(24.). The affection of B is ambiguous, unless it can be determined by this rule, that according as AC+BC is greater or less than 180°, A+Bis greater or less than 180°, (10.). From ACB the angle sought draw CD perpendicular to AB; then R: cos AC:: tan A: cot ACD, (c. 3.); and tan BC: tan AC: cos ACD: cos BCD, (28.)ACD +BCD ACB, and ACB is ambiguous, because of the ambiguous sign+ or

=

Let fall the perpendicular CD
from the angle C, contained by
the given sides, upon the side
AB. R: cos A:: tan AC: tan
AD, (c. 2.); cos AC: cos BC::
cos AD: cos BD, (26.).
AB=AD+BD, wherefore AB is
ambiguous.

TABLE continued.

GIVEN.

Two angles

A, B,

and a side

AC opposite to

one of them,

B.

The three

sides,

AB, AC,

and

BC.

SOUGHT.

The side
BC
opposite

to the
other
given an-
gle A.

The side
AB
adjacent
to the
given
angles

A, B.

The third

angle

ACB.

One of the

angles

A.

SOLUTION.

Sin B: sin A:: sin AC: sin BC, (24.); the affection of BC is uncertain, except when it can be determined by this rule, that according as A+B is greater or less than 180°, AC+BC is also greater or less than 180°, (10.).

From the unknown angle C, draw CD perpendicular to AB; then R: cos A tan AC: tan AD, (c. 2.); tan B: tan A :: sin AD: sin BD. BD is ambiguous; and therefore AB = AD + BD may have four values, some of which will be excluded by this condition, that AB must be less than 180°. From the angle required, C, draw CD perpendicular to AB.

R: cos AC: tan A: cot ACD, (c. 3.), cos A cos B:: sin ACD: sin BCD, (25.). The affection of BCD is uncertain, and therefore ACB = ACD + BCD, has four values, some of which be excluded by the condition, that ACB is less than 180°.

may

From C one of the angles not required, draw CD perpendicular to AB. Find an arch E such that tan AB: tan (AC+BC):: tan

(AC-BC): tan E; then, if AB be greater than E, AB is the sum, and E the difference of AD and DB; but if AB be less than E, E is the sum, and AB the difference of AD, DB, (29.). In either case, AD and DB are known, and tan AC tan AD :: R: cos A.

11

10

In the foregoing table, the rules are given for ascertaining the affection of the arch or angle found, whenever it can be done: Most of these rules are contained in this one rule, which is of general application, viz. that when the thing found is either a tangent or a cosine, and of the tangents or cosines employed in the computation of it, either one or three belong to obtuse angles, the angle found is also obtuse. This rule is particularly to be attended to in cases 5. and 7. where it removes part of the ambiguity.

It may be necessary to remark with respect to the 11th case, that the segments of the base computed there are those cut off by the nearest perpendicular; and also, that when the sum of the sides is less than 180°, the least segment is adjacent to the least side of the triangle: otherwise to the greatest (17.).