In this case, as in the problems of the second class, the only requirement here is to find a second angle; for then the problem becomes one belonging to the first class. But the finding of the second angle, in problems of the third class, is attended with more computation than it is in problems of the second class. The process is as follows: Having one angle of a triangle, the sum of the two remaining angles is obtained by subtracting the given angle from 180o - the sum of the three angles.
Then with equation (203.) the difference of the two angles is obtained. And then, having the sum and difference of the two angles, either may be found by one of the equations (204.) and (205.).
For example: let Fig. 320 represent the triangle in which a (= 36 feet) and b(= 27 feet) are the given sides;
and C (= 105o) the angle included between the given sides, a and b. The sum of the two angles A and B, therefore, will be -
(A +B)= 180- 105 = 75°,
and the half of the sum of A and B is 75/2 = 37° 3°'.
The sum of the given sides is 36 + 27 = 63, and their difference is 36 - 27 = 9.
Then from equation (203.) we have -
. tan. 1/2(A - B)= tan. 37o 30 9/63. Solving this by logs. (Art. 427), we have -
Log. tan. 370 30'
tan..1/2(A-B)(=6° 15' 20.5")
Thus half the difference of A and B is 6° 15' 20. 5", nearly,
By equation (204.) -
6° 15' 20.5"
31° 14' 39.5"
and by equation (205.) -
The given angle,
105. 0. 0
The three angles,
180. 0. 0
Thus, by adding together the three angles, the work is tested and proved.
Having the three angles, the third side may now be found by the rule for problems of the first class.
481. - Oblique-Angled Triangles: Fourth Class. - The fourth class comprises those problems in which the three sides of the triangle are given, to find the three angles.
The method by which the problems of the fourth class are solved is to divide the triangle into two right-angled triangles; then, by the use of equation (129.), to find one side of one of these triangles, and then with this side to find one of the angles, then by rules for the second class problems, obtain the second and third angles.
Thus, from equation (129.), we have -
g = c2-(a+b) (a-b)
By the relation of sines to sides (Art. 476), we have (Fig. 321) -
b: g:: sin. E: sin. F.
But the angle E is a right angle, of which the sine is unity, therefore -
b: g 1: 1: sin. F =g/b.
Substituting for g its value as above, we have -
sin. F =c2-(a+b) (a-b)/2bc. (200.)
To illustrate: let a, b. c (Fig. 321) be the three given sides of the triangle ABC, respectively equal to 12, 8 and 16 feet. With these, equation (206.) becomes -
Sin. F = 162 - (12+8) (12-8)
2 x 8 x 16 sin.F = 256 - (20 x 4)
sin. F = 176.
Solving this by logarithms (Art. 427), we have -
Log. sin. 43o 26'
or, the angle at F equals 43o 26', nearly. Of the triangle
A C E (Fig. 321), E is a right angle, therefore the sum of F and A, the two remaining angles, equals 900 (Art. 346). Hence, for the angle at A, we have -
A = 90o - 43° 26' = 46o 34'.
We now have two sides a and b and A, an angle opposite to one of them, to find B, a second angle. For this, equation (199.) is appropriate. Thus -
sin. B = sin. A b/a.
This may be solved as shown in Art. 478.
And, when the second angle is obtained, the third angle is found by subtracting the sum of the first and second angles from 180o.
But to test the accuracy of the work, it is well to compute the angle C from the angle A, and the sides a and c. For this, equation (201.) will be appropriate.
482. - Trigonometric Formulae: Right-Angled Triangles. - For facility of reference the formulae of previous articles are here presented in tabular form. The symbols referred to are those of Fig. 322. formulae in tabular form. 531
B= 90° -A.
A = 90° - B.
b = a tan. B.
c =a/cos. B
a = b/tan. B
sin.= b/ sin. B
a = c cos. B.
b = c sin. B.
483. - Trigonometrical Formulae: First Class, Oblique.
__The symbols of the formulae of the following table indicate quantities represented in Fig. 323 by like symbols.
A, B, b,
a = b sin. A/sin. B.
A, C, c,
a = c sin. A/sin. C.
A, B, a,
b = a sin. B/sin. A.
B, C, c,
b = c sin. B/sin. C.
A, C, a,
c = a sin. C./sin. A.
B, C, b,
c = b sin. C/sin. B.
484. - Trigonometrical Formulae: Second Class, Oblique. - The symbols in the formulae of the following table refer to quantities represented in Fig. 323, by like symbols.
B, a, b,
sin. A = sin. B a/b.
C, a, c,
sin. A = sin. C a/c.
A, a, b,
sin. B = sin. A b/a.
C, b, c,
sin. B = sin. C b/c.
A, a, c,
sin. C = sin. A c/a
B, b, c,
sin. C = sin. B c/b.
See Formulae, First Class.
485. - Trigonometrical Formulae: Third Class, Oblique. - The symbols in the formulas of the following table refer to quantities shown by like symbols in Fig, 323.
C, a, b,
tan. 1/2(A - B) = tan. 1/2(A+B)a-b/a+b.
A=1/2(A+B)+ 1/2(A -B).
B =1/2(A +B) - 1/2(A - B).
C + B,
C +B= 180 - A.
A, b, c,
tan. 1/2(C - B) = tan. 1/2 (C + B) c-b/c+b,
C = 1/2(C + B) +1/2(C-B).
B=1/2(C + B)-1/2(C-B).
C + A.
C + A = 180 - B.
B, a, c,
tan. 1/2(C-A) = tan. 1/2(C + A) c-a/c+a.
C=1/2(C + A) + 1/2(C-A).
A=1/2(C + A) + 1/2(C-A).
For the remaining side consult formulas for the first class.
486. - Trigonometrical Formulae: Fourth Class, Oblique. - The symbols in the formulas of the following table refer to quantities shown by like symbols in Fig. 321.
Given a, b, c, to find A, B, C.
sin. F =c2 - (a+b) (a-b)
A = 90 - F.
sin. B = sin. A b/a.
sin. C = sin. A c/a.
C =180 - (A + B).