Log. AC = log. R + log. BC - log. sine A.

Log. R + log. 67

Log. sine 51° 39′ 15′′

11-8260748

9-8944713

1-9316035

Log AC = 85-42834

2

Or, since AC = AB + BC2 (47.1), AC = √ 7298 =

85-428+.

Also, AC may be found more easily by the secant. R: sec. ∠A :: AB : AC.

When a proportion can be formed, in which radius is the first term, the operation becomes more easy. Thus, the operation by the last analogy, where radius is the first term, will be more easy than that by the former analogy, where the sine of A is the first term.

Ex. 2. Given the base AB = 20, and the perpendicular BC = 30, to find the angles and the hypothenuse. Answer. Angle A = 56°18′, C = 33° 42′, AC = 36.06.

86. Remark. In some instances the tangents and secants are more adapted to produce accurate results than the sines and cosines. The reason of the deficiency in the sines and cosines is their small variation in certain parts of their arcs. Thus, if an arc near 90o be found in terms of its sine, and if a very small arc, or an arc near 180°, be found in terms of its cosine; the variations of the sines and cosines of these arcs are so small, that the sines and cosines will not change in the tables for many seconds of the arcs. For example, if the log. sine, or log. cosine of an arc should happen to be 9.9999998, this log. in the tables, is the sine of an arc from 89° 56′ 19′′ to 89° 57' 8", or the cosine of an arc from 2' 52" to 3' 41". Consequently it is impossible to know what arc or angle, between these limits, is to be taken. The reason of this uncertainty is, that the log. sines and log. cosines, in the tables, are not continued to more than seven places of decimals, and in some tables only to five or six decimals.

In these cases it will be expedient to employ the log. tangents, or log. cotangents, which are not liable to this defect, as the difference for 1" is 42 at an arc of 45°, and greater in every other part of the quadrant.

But when a sine or cosine of this kind enters into a calculation, as one of the data, it is rather favourable than otherwise to the accuracy of the result, because any small error in the quantity of the given arc or angle will not affect the tabular value of its sine or cosine.

Again, if a required arc or angle be either very small, or very near 180°, we may express it by a sine instead of a cosine, as follows.

Because AF =

2 AL2
CA

2 AL2

CA (46. no. 3), CF=CA-AF=

CA- → that is, cos. ACB=r

2 sine2 ACB

ACB

21
2

r

, therefore sine A =

A.

r

Let A denote any

; hence sine A

√(r2-rx cos. A)

2

•

Now as the log. sine of a small arc increases fast, we shall thus obtain the arc or angle A true to a second.

In like manner, if an arc very near 90o be expressed by its sine, it can be known only within certain limits, by reason of the small variation of the sine. We must therefore express the arc in terms of the cosine, which may be done by means of the

formula, sine A

r-rx cos. A

2

70. Solution of the Cases of Oblique-angled Triangles.
Fig. page 18.

Case 1. Given the angle A 49° 25′, the angle C 63° 48′, and the side AB 275, to find the rest.

The angle B = 180° - (49° 25′ + 63° 48′) = 66° 47'. Draw AB = 275 from a scale of equal parts. By a scale of chords make the angle A = 49° 25', and the angle B = 66° 47'. Draw AC and BC meeting each other in C. Then AC and BC measured by the scale of equal parts will be 281·6 and 232.7.

Sine C: sine A :: AB : BC.

Sine 63° 48′: sine 49' 25' :: 275 : BC.

Log. sine A + log. AB-log. sine C = log. BC.

Sine 63" 48′ : sine 66° 47' :: 275 : AC.

Ex. 2. Given the angle A 46° 22', the angle B 79° 23', and the side AC 135 feet, to find the rest. Answer. The angle C = 54° 15', the side AB = 111·47 feet, and BC = 99.41.

Case 2. Given the side AB 532, BC 358, and the angle C 107° 40', to find the rest.

Draw BC=358 from a scale of equal parts. By a scale of chords make the angle C = 107° 40′. From the centre B, with the radius BA = 532, intersect CA in A. Then AC measured by the scale of equal parts is 299.6, and the angles A and B measured by the scale of chords are 39° 53′ and 32° 27'.

As the angle C is obtuse, the radius BA cannot intersect CA in more points than one; therefore there cannot be two triangles; consequently this example is not ambiguous (75).

As the angle C is obtuse, each of the angles A and B must be acute, or less than 90; therefore this example is not ambiguous (75). Hence the angle B = 180° - (39 53' + 107 40') = 32° 27′.

Sine C: sine B :: AB : AC.

Sine 107 40': sine 32 27':: 532: AC.

Ex. 2. Given the side AC = 20, BC = 30, and the angle A = 56 14', to find the rest.

Answer. The angle B = 33° 39′, C = 90° 7', and the side AB = 36.088.

Ex. 3. Given AC 236, BC 350, and the angle B 38 40', to find the rest.

Draw BC = 350, and make the angle B = 38° 40′. From the centre C, with the radius CA = 236, describe an arc, which will cut BA in two points. From the two points of intersection of BA and the radius CA draw lines to C. Then there will be two triangles, each of which will give the parts required. Hence this example is ambiguous (75). The angle C is 29 or 73, the angle A is 67 or 112', the side AB is 184 or 362.

Case 3. Given the side AB 176, BC 133, and the included angle B 73, to find the rest.

AB + BC : AB - BC :: tan. (A+C): tan. (C-A).

53°30′: tan.

(C-A).

Ex. 2. Given one side 40, another 70, and the inclu

ded angle 76° 14', to find the rest.

Answer. The other two angles are 71° 3 and 32° 43′, and the third side is 71.88.

Case 4. Given the three sides, AB 350, AC 240, BC 200, to find the angles. See fig. page 16.

AB: AC + BC :: AC - BC: AD - DB = AG. 350: 440 :: 40 : 50.29.

AD = 1⁄2 (AB + AG) = 175 + 25.145 = 200.145. DB = (AB-AG) = 175 - 25.145 = 149.855. AC: AD :: R: sine ACD.

240: 200.145 :: R: sine ACD = 56° 30′.

BC: BD :: R : sine BCD.

200: 149-855 :: R: sine BCD = 48° 32′.

Hence the angle A = 33° 30', B = 41° 28′, and ACB = 105° 2'.

Ex. 2. Given the three sides, AB = 60, AC = 50, BC = 40, to find the angles.

Answer. The angle B = 55° 46', A = 41°24′, ACB = 82° 50′.