11.

I.

two branches, is resolved by the application of propositions 9 and 10; and the solution of the third case flows immediately from the former of these propositions.

SOLUTION.

1

AB. BC: (P-AB) (¿P-BC) :: R': sinB
P(P-AC): (P-AB) (P-BC):: R2; tan B2. 2
P (P-AC) :: R2: cos B2. 3

:

AB.BC:
2AB. BC: AB2 + BC2 -AC2 : : R cosB.

4

AB: BC sinC; sin A; whence B, and
sinC:sinB:: AB: AC.

5

6

AB+BC: AB-BC:: cotB;; cot(A+B).

AB: BC::R:: tan b; and

7

or - cot (C+B).

Į R: tan(45°-b) :: cotžВ : cot (A+†B),

8

and

AC.

or - cot (C+B).

sin A: sinB:: BC: AC, or

AC=√(AB+BC2-2AB.BC cosB.)

9

10

For the resolution of the first Case, the analogy set down first, is on the whole the most convenient, particularly if the angle sought do not approach to two right angles. The second analogy may be applied with obvious advantage through the entire extent of angles. The third and fourth analogies, especially the latter, are not adapted for the calculation of very acute angles; they will, however, answer the best when the angle sought is obtuse. It is to be observed, that the cosines of an angle and of its supplement are the same, only placed in opposite directions; and hence the second term of the analogy, or the difference of AB2+BC2 from AC2, is in excess or defect, according as the angle at B is acute or obtuse.-These remarks are founded on the unequal variation of the sine and tangent, corresponding to the uniform increase of an arc.

The first part of Case II. is ambiguous, for an arc and its supplement have the same sine. This ambiguity, however, is removed if the character of the triangle, as acute or obtuse, be previously known.

For the solution of the second part of Case II. the first analogy is the most usual, but the double analogy is the best adapted for logarithms. In astronomy, this mode of calculation is particularly commodious. The direct expression for the side subtending the given angle is very convenient, where logarithms are not employed.

PROP. XVII. PROB.

Given the horizontal distance of an object and its angle of elevation, to find its height and absolute distance.

Let the angle ABC, which an object A makes at the station B, with an horizontal line, and also the distance BC of a perpendicular AC, to find that perpendicular, and the hypotenusal or aërial distance BA.

In the right-angled triangle BCA, the radius is to the tan

B

gent of the angle at B, as BC to AC; and the radius is to the secant of the angle at B, or the cosine of the angle at B is to the radius, as BC to AB.

PROP, XVIII. PROB.

Given the acclivity of a line, to find its corresponding vertical and horizontal length.

In the preceding figure, the angle CBA and the hypotenusal distance BA being given to find the height and the horizontal distance of the extremity A.

The triangle BCA being right angled, the radius is to the sine of the angle CBA as BA to AC, and the radius is to the cosine of CBA as BA to BC.

Scholium. If the acclivity be small, and A denote the

measure of that angle in minutes; then AC-BA×

A 3438

nearly. But the expression for AC, will be rendered more accurate, by subtracting from it, as thus found, the

In most cases when CBA is a small angle, the horizontal distance may be computed with sufficient exactness, by de

ducting

AC2
2BA'

or BAXA X.000,000,0423, from the hy

potenusal distance.

PROP. XIX. PROB.

Given the interval between two stations, and the direction of an object viewed from them, to find its distance from each.

Let BC be given, with the angles ABC and ACB, to calculate AB and AC.

In the triangle CBA, the angles ABC and ACB being given, the remaining or supplemental angle BAC is thence given; and consequently, sin BAC: sinACB:: BC: AB, and sinBAC: sin ABC:: BC: AC.

Cor. If the observed angles ABC and ACB be each of them 60°, the triangle will

be evidently equilateral; and if the angle at

B

the station B be right, and that at C half a right angle, the distance AB will be equal to the base BC.

PROP. XX. PROB.

Given the distances of two objects from any station and the angle which they subtend, to find their mutual distance.

Let AC, BC, and the angle ACB be given, to determine AB.

In the triangle ABC, since two sides and their contained angle are gi ven, therefore, by corollary to Proposition 10. AC + BC: AC BC :: cot C: cot(A+C), then sinA : sinC : : BC: AB; or (from the cor. to Prop. 11.) AB=√(AC2+BC-2AC.BC cosC.)

Α.

B

Cor. By combining this with the preceding proposition, the distance of an object may be found from two stations, between which the communication is in

terrupted. Thus let A be visible from B and C, though the straight line BC cannot be traced. Assume a third station D, from which B and C are both Measure DB and DC, and observe the angles BDC, ABC and ACB. In the triangle BDC, the base BC is found as above; and thence, by the preceding proposition, the sides AB and AC of the triangle ABC are determined,

seen.

B

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