408 
APPENDIX A. 
Fig. 415. 
B 
12. Oblique-angled Triangles. Let ABO be any oblique-angled tri 
angle, tbe angles and sides being noted as in the figure. Then any three of 
its six parts being given, and one of them 
being a side, the other parts can be ob 
tained by one of the following methods, 
which are founded on these three theo 
rems : 
Tiieokem I.—In every plane triangle, 
the sines of the angles are to each other as 
the opposite sides. 
Theorem II.—In every plane triangle, the sum of two sides is to their dif 
ference as the tangent of half the sum of the angles opposite those sides is to 
the tangent of half their difference. 
Theorem III.—In every plane triangle, the cosine of any angle is equal to 
a fraction whose numerator is the sum of the squares of the sides adjacent to 
the angle, minus the square of the side opposite to the angle, and whose de 
nominator is twice the product of the sides adjacent to the angle. 
All the cases for solution which can occur may be reduced to four: 
Case 1.— Given aside and two angles. The third angle is obtained by 
subtracting the sum of the two given angles from 180°. Then either un 
known side can be obtained by Theorem I. 
Calling the given side a. we have b — a. ^4——- ; and c = a — 
sm. A sin. A 
Case 2.— Given two sides and an angle opposite one of them. The angle 
opposite the other given side is found by Theorem I. The third angle is ob 
tained by subtracting tbe sum of the other two from 180°. The remaining 
side is then obtained by Theorem I. 
Calling the given sides a and b, and the given angle A, we have sin. B = 
I 
sin. A . —. 
a 
Since an angle and its supplement have the same sine, the result is am 
biguous; for the angle B may have either of the two supplementary values 
indicated by the sine, if 5 > a, and A is an acute angle. 
C = 180° - (A + B). c = sin. C . ° - . 
sin. A 
Case 3.—Given two sides and their included angle. Applying Theorem 
II (obtaining the sum of the angles opposite the given sides by subtracting 
the given included angle from 180°), we obtain the difference of the unknown 
angles. Adding this to their sum we obtain the greater angle, and subtract 
ing it from their sum we get the less. Then Theorem I will give the remain 
ing side. 
Calling the given sides a and b, and the included angle C, we have 
A + B = 180° - C. Then 
¿(A—B) =tan. HA + B) a ~ 1) 
tan. 
a + b 
*(A + B) + $(A - B) = A. i ( A + B) — i (A - B) = B. e = a 
sin C. 
sin. A'