Ex. 1. Given a = 4602.21059, c = 4602.836; find B.

NOTE. -The characteristic 5 is increased numerically to 6 to make it divisible by 2 (see Note 4 of Art. 66). Ten is then added to the characteristic 3, making it 7, so as to agree with the Tables (Art. 76).

There is a slight error in the above value of B on account of the irregular differences of the log sines for angles near 0° (Art. 81). A more accurate value may be found by the principle that the sines of small angles are approximately proportional to the angles (Art. 130).

EXAMPLES.

The following right triangles must be solved by logarithms.

1. Given a = 60, c = 100; find A, B, b.

Ans. A 36° 52'; B = 53° 8'; b = 80.

2. Given a = 137.66, c = 240; find A, B, b.

Ans. A = 35°; B = 55°; b 196.59.

Ans. B = 50°; a= 64.279; b=76.604.

6. Given A = 30°, c= 150; find B, a, b.

7. Given A= 32°, c = 1760; find B, a, b.

75; b=75√3.

Ans. B = 58°; a = 932.66; b=1492.57.

8. Given A 35° 16′ 25", c = 672.3412; find B, a, b.

9. Given A

=

Ans. B 54° 43′ 35′′;

=

a = 388.26; b=548.9.

Ans. B 15°; b= 80 (2-√3); c=

=

=80(√6-√2).

715.72; c= 884.68.

10. Given A= 36°, a = 520; find B, b, c.

11. Given A = 34° 15', a = 843.2; find B, b, c.

Ans. B = 55° 45'; c=1498.2.

Ans. B = 22° 22′ 45′′; a = 618.66; c=669.05.

12. Given A = 67° 37' 15", b =

254.73; find B, a, c.

13. Given a =

75, b=75; find A, B, c.

=

=

Áns. A =

46° 23' 50"; c=

= 29.

Ans. A = 45° B; c=75√2.

14. Given a 21, b = 20; find A, B, c.

15. Given a = 300.43, b = 500; find A, B, c.

Ans. A = 31°; B = 59°; c = 583.31.

16. Given a = : 4845, b = 4742; find A, B, c.

OBLIQUE TRIANGLES.

114. There are Four Cases of Oblique Triangles.

I. Given a side and two angles.

II. Given two sides and the angle opposite one of them. III. Given two sides and the included angle.

IV. Given the three sides.

The formulæ for the solution of oblique triangles will be taken from Chap. VI. Special attention must be given to the following three, proved in Arts. 95, 96, 97.

115. Case I. Given a side and two angles, as a, B, C; find

Ex. 1. Given a = 7012.6, B = 38° 12′ 48′′, C = 60°; find

Ex. 2. Given a=1000, B=45°, C = 127° 19'; find A, b, c. Ans. A = 7° 41'; b = 5288.8; c = 5948.5.

=

Given two sides and the angle opposite one A; find B, C, c.

b sin A

116. Case II.

of them, as a, b,

(1) sin B:

a

; thus B is found.

C 180° (A+B); thus C is found.

This is usually known as the ambiguous case, as shown in geometry (B. II., Prop. 31). The ambiguity is found in the equation

Since the angle is determined by its sine, it admits of two values, which are supplements of each other (Art. 29). Therefore, either value of B may be taken, unless excluded by the conditions of the problem.

I. If absin A, sin B >1, which is impossible; and therefore there is no triangle with the given parts.

II. If a = b sin A, sin B = 1, and B = 90°; therefore there is one triangle-a right triangle-with the given parts.

III. If a >b sin A, and <b, sin B <1; hence there are two values of B, one being the supplement of the other, i.e., one acute, the other obtuse, and both are admissible; therefore there are two triangles with the given parts.

IV. If a >b, then A > B, and since A is given, B must be acute; thus there is only one triangle with the given parts.

These four cases may be illustrated geometrically.

Draw A, the given angle. Make AC = b; draw the perpendicular CD, which = b sin

III. If a>b sin A, and <b, the circle cuts AX in two points B and B', on the same side of A; thus there are two triangles ABC and AB'C, each having the given parts, the angles ABC, AB'C being supplementary.

Β ́ A

D

B

IV. If a>b, the circle cuts AX on opposite sides of A, and only the triangle ABC has the given parts, because the angle B'AC of the triangle AB'C is not the given angle A, but its supplement.

These results may be stated as follows: