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arc corresponding to BC. But the radius is equal to an sp g eq

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the arc which corresponds to BC, as 3 ×57, or 172°, to the number of degrees contained in the angle CAB, and consequently 2AC+AB: BC: ; 172°: the expression of the angle at A, or AC+}AB : BC :: 86°: number of degrees in the angle at A. This approximation will be the more correct, when the side opposite to the required angle becomes small in comparison with the hypotenuse; but the quantity of error can never amount to 4 minutes.


Three variable parts of an oblique angled triangle being given, to find the other two.

This general problem includes three distinct cases, one of which again is branched into two subordinate divisions.

1. When all the three sides are given.

2. When two sides and an angle are given ; which angle may either (1.) be contained by these sides, or (2.) subtended by one of them.

3. When a side and two of the angles are given.

The first case admits of four different solutions, derived from Propositions 11, 12, 13, and 14, and which have their several advantages. The second case, consisting of 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.

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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 AB^+BC” from AC", 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.



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 hy- -
potenusal or aerial distance B.A. -
: In the right-angled triangle _T

BCA, the radius is to the tan- * —. 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.


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 f, 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=BAx _A_


nearly. But the expression for AC, will be rendered

more accurate, by subtracting from it, as thus found, the 3.

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In most cases when CBA is a small angle, the horizontal . .

distance may be computed with sufficient exactness, by deAC” 2BA potenusal distance,


, or BAx A* x.000,000,0423, from the hy


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



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