By def. 27. sect. 4. The sine of 96° 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°.

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 36o 40a 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 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

36°. 40143°. 20/ the sum of the angles Cand B: therefore half of 143°. 20', will be half the sum of the two required angles, C and B.

By theo. 2. of this sect.

As the sum of the two sides AB and AC = 420

is to their difference,

So is the tangent of half the sum of

the two unknown angles C and B

}

=

71° 404

to the tangent of half their difference = 23° 20

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 C and B being found; BC is had, as before, by theo. 1. of this sect. Thus,

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 triangle into two right angled ones, ADC and CBD; as well as the base AB, into the two segments, AD and DB.

is to the sum of the other sides, AC and BC, 81

To half the base, or to half the sum

32

Add half their difference, now found,

8.23

of the segments AD and DB.

Their sum will be the greatest segment AD 40.23

S

In the right angled triangle ADC, there is AC 47, and AD 40. 23, given, to find the angle A.

This is resolved by case 4. of right angled plane trigonometry, thus,

AD: R:: AC: Sec. A.

40.23: 90°:: 47: 31o 08′′

Or it may be had by finding the angle ACD, the complement of the angle A; without a secant, thus,

AC: R::AD: S. ACD.

44 900 40. 23: 58° 52′

90-58° 52/=319. 08/, the angle A.

Then by theo. 1 of this sect.

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

34: 3108: : 47: 45° 37.

By cor. 1. theo. 5. sect. 4. 180°-the sum of A and BC.

A 31.08!

B 45. 37

180°-76. 45-103°. 15', the angle C.