If the side opposite the given angle be equal to, or greater than the other given side, or the given angle obtuse; then there would be but one answer to the problem, because the angle, opposite that other given side, will be always acute; but when the given angle is acute and opposite the lesser of the given sides, the answer is ambiguous, as the sine of an angle is equal to the sine of its supplement, consequently the required angle opposite that other given side may be obtuse, or acute; unless it is given in the conditions of the problem.

In the last problem the given angle is acute, and the side opposite to it less than the other given side, therefore the angle C may be acute or obtuse; but the side and angle answering to the acute value of C has been already found. Now it remains to find the side and angle of the triangle, answering to the obtuse value of C, which is thus found:

The acute value of C, found in the foregoing calculation, is 60° 31', consequently its obtuse value is 180° 60° 31'=119° 29', then 119° 29′+46° 30′, taken from 180°, gives 14° 1' to the remaining angle ABC. (Pl. 14. Fig. 5.)

To find the side AC, answering to the obtuse value of the angle C.

As the sine of A=

CASE II.

Two angles and a side given; to find the other sides.

PL. 5. fig. 15.

In the triangle ABC, there is the angle A 46° 30′, AB 230, and the angle B 37o 30', given to find AC and BC.

1st. By Construction.

Draw a blank line, upon which set AB 230, from a scale of equal parts; at the point A of the line AB, make an angle of 46° 30′, by a blank line; and at the point B of the line AB make an angle of 37° 30', by another blank line: the intersection of those lines gives the point C, then the triangle ABC is constructed. Measure AC and BC from the same scale of equal parts that AB was taken; and you have the answer required.

2d. By Calculation.

By (cor. 1. theo. 5. sect 4.) 180°-the sum of the angles A and B =C,

A 46° 30′

B 37. 30

180°-84°. 00′=96° 00′=C.

By def. 27. sect. 4. The sine of 96o-the sine of 84°, which is the supplement thereof; therefore instead of the sine of 96°, look in the tables for the sine of 84°.

By theo. 1. of this sect.

Extend from 84° (which is the supplement of 96) to 46° 30′ on the sines; that distance will reach from 230 to 168, on the line of numbers, for BC.

Extend from 84° to 37°. 30', on the sines; that extent will reach from 230 to 141, on the line of numbers, for AC.

CASE III.

Two sides and a contained angle given; to find the other angles and

side.

PL. 5. fig 16.

In the triangle ABC, there is AB 240, the angle A 36° 40′ and AC 180 given; to find the angles C and B, and the side BC.

1st. By Construction.

Draw a blank line, on which from a scale of equal parts, lay AB 240; at the point A of the line AB, make an angle of 36° 40′, by a blank

line; on which from A, lay AC 180, from the same scale of equal parts; measure the angles C and B, and the side BC, as before; and you have the answers required.

2d. By Calculation

By cor. 1. theo. 5. sect. 4. 180°-the angle A 36o. 40 143°. 20', the sum of the angles C and B: therefore half of 143°. 20', will be half the sum of the two required angles, C and B.

By then. 2. of this sect.

As the sum of the two sides AB and AC≈ 420 is to their difference,

60.

}

=71° 40′

23°20′

So is the tangent of half the sum of
the two unknown angles C and B
to the tangent of half their difference

By theo. 4.

To half the sum of the angles Cand B=71° 40' Add half their difference as now found = 23 20

The sum is the greatest angle, or ang. C-95 00

Subtract, and you have the least angle, or B=48 20

The angle Cand B being found; BC is had as before, by theo. 1. of this sect. Thus,

S. B: AC:: S: A: BC

48° 20': 180:: 36° 40': 143, 9.

3d. By Gunter's Scale.

Because the two first terms are of the same kind, extend from 420 to 60 on the line of numbers; lay that extent from 45° on the line of tangents, and keeping the left leg of your compasses fixed, move the right leg to 71°. 40'; that distance laid from 45° on the same line will reach to 23°. 30', the half difference of the required angles. Whence the angles are obtained, as before.

The second proportion may be easily extended, from what has been already said.

CASE IV.

PL. 5. fig. 17.

The three sides given, to find the angles.

In the triangle ABC, there is given, AB 64, AC 47, BC 34: the angles A, B, C, are required.

1st. By Construction.

The construction of this triangle must be manifest, from prob. 1. sect. 4.

2d. By Calculation.

From the point C, let fall the perpendicular CD on the base AB; and it will divide the tringle into two right angled ones, ADC and CBĎ; as well as the base AB, into the two segments, AD and DB.

AC 47
BC 34

Sum 81

Difference 13