the other figures, index and all, from 9, setting down the remainders, and piace a dot before the index, as in the case of the logarithm.Thus the arithmetical complement (usualy marked Co. Ar.) of the logarithm 9.66004 is 0·33996, and so of any other.

When the arithmetical complement of the first term is used instead of the logarithm, add all the three numbers together, and reject 10 out of the index of their sum, as in those cases where the radius is the first term.

PROBLEM I. In the olique angled triangle ABC, given two angles and a side opposite to one of them, to find the two other sides.

Suppose the angle at A 36°,40', the angle at B, 60°,51', and the base AB 85.6: Required AC and BC?

Geometrically. Draw the base AB, and from any scale of equal parts, lay thereon 85.6 from A to B; then, from the line of chords, lay off an angle of 36°,40′ at A, and an angle of 60°,51' at B, and the meeting of these two lines in C completes the triangle, and AB A. and BC may be measured by the same line of equal parts.

From the sum of all the angles

180°,00'

Take the sum of the angles A and B, viz. 97°,31′

And we have the angle C equal to

82°,29'

Here we have the angle at C opposite to the given base, and the angles at A and B opposite the two required sides, which may be found by the first rule, as follows:

By Calculation. For the side BC. Having to find a side, we begin with an angle,

PROBLEM II. and III. Two sides and an angle opposite to one of them, giv. en, to find the two other angles and remaining side.

In the oblique angled triangle ABC, given the side AC 75.4, the side BC 51-56, and the angle at A 369,40', to find the base AB, and the angles at B and C.

Geometrically.

Geometrically. Draw the base AB, at pleasure, and on any point assumed, as A, make an angle of 36°,40'; take 75.4 from the scale of equal parts and set it from A to C, then take 51.56 from the same scale; set one foot of the dividers in C, and with the other intersect the base in B; lastly, draw BC, and the triangle is completed, and the base may be measured by the same scale of equal parts.

B

By Calculation. Here we have the side BC opposite the known an gle at A, and the side AC opposite the unknown angle at B, which may be found by Rule 1st.

For the base AB. Having to find a side we begin with an angle. 36°,40' Co. Ar. 0.22392

As the sine of A

To find the tangent of half the difference of two unknown angles by the scale.

RULE 1.-When the tangent of half the sum of the unknown angles is less than 45°, the extent from the half sum of the sides to their half difference on the line of numbers will reach from the half sum of the unknown angles to their half difference on the tangents.

2. When the tangent of half the sum of the unknown angles exceeds 45°, take the extent between the sum and difference of the sides; set one foot in tangent's radius, 45°, and fix the other foot wherever it falls on the tangent line, and contract the foot that stands on 45°, to the tangent of the half sum of the unknown angles; then with that extent set one foot in 45°, of tangents, and the other will point out the tangent of the half difference.

PROBLEM IV. and V. Two sides, and the angle included between them at A, given, to find the two other angles and the other side.

In the oblique angled triangle ABC, given the side AC 75-4, the base AB 85.6, and the included angle at A 36°,40′, to find 'the angle B and C, and the side BC.

Geometrically.

Geometrically. Draw the base AB, and from any scale of equal parts, set off 85.6 from A to B ; make an angle at A of 36°, 40'. and draw AC, and from the same scale of equal parts, set 75.4 from A to C ; lastly, draw the line BC, and the triangle is completed BC may be measured by the

B

same scale of equal parts, and the angles B and C, on the line of chords.

By Calculation. Here we have given the two sides AB and AC, with the angle included between them; and therefore these cases must be solved by Rules 2d and 1st. Now, as the three angles of every triangle are equal to 180°, the angle at A 36°,40' being subtracted from 180° leaves 143°,20', the sum of the two unknown angles B and C, half of which is 71°,40'; and half their difference may be found by the following proportion, according to Rule 2.

As the sum of the two sides AB and AC 161 Co. Ar. 7.79318
Is to their difference

So is the tangent of half the sum of
the unknown angles B and C

To the tangent of half their difference

1.00860

10.2

}

710,40'
10°,49'

10-47969

9.28147

71°,40'

10,49

From the half sum
Take the half diff.

71°,40 10,49

The sum is the greater ang. C 82,29

Having found the angles B and C,

the side BC may be found by

The remainder is
the less angle B

}

60,51

To the half sum

Add the half difference

Rule 1.

PROBLEM VI. The three sides given, to find the angles.

In the oblique angled triangle ABC, given the base AB 85.6, the side AC 75-4, and the side BC 51-56; Required the angles?

Geometrically. Draw the base AB, and set off 85'6 from any scale of equal parts from A to B; take 75.4 from the same scale, and setting one foot in A, describe an arch; then from the scale take 51.56, and setting one foot in B, intersect the former arch in C; from C draw lines to A and B, and the triangle is completed. The angles may all be measured upon the line of chords.

E

D

be By Calculation. Here being no angle given, these cases must solved by Rule 3d, in the following manner: Place one foot of the dividers in C, and extend the other so as to take in the shortest side BC, and describe the arch BE; then, from C let fall a perpendicular on the base AB, which will divide it into two segments, AD the greater, and DB the less, whose difference is AE: Then, 3...B

A's

As the Base AB

85-6

Is to the sum of the two sides AC and BC 126.96
So is the difference of the sides AC and BC 23.84
To the difference of the segments

of the base, or AE

}

35.36

17.68

42.8 From half the base

42.8

17.88) Take the half difference 17.68

60.48

}

And the remainder is
the less segment DB

Thus is the oblique angled triangle ABC divided into two right angled triangles ADC and BDC, both right angled at D, in each of which are given the base and hypothenuse, to find the other parts.

First, For the angle at C in the right angled triangle ADC, making the hypothenuse radius.

Co. Ar. 8.12263

60.48

1.78161

90°,00'

10.

To the sine of C

53°,20'

9.90424

The angle A, being the

complement of the angle C, is 36o,40'.

Then for the angle C in the right angled triangle BDC.

Whence the angle B is 60°,51', being the complement of 29°,09'; and the angle at C, in one triangle, being added to the angle C in the other, is 82°,29'; thus the solution of the problem is finished.

Trigonometry is easily applied to Navigation, and the Mensuration of Heights and Distances. With respect to the former; suppose in the first Problem of right angled Trigonometry, the angle at A is the ship's course, the base to be the true, (or meridional,) difference of latitude, the perpendicular to be the departure, or difference of longitude, and the hypothenuse to be the distance the ship is to run; then we have the course and true (or meridional) difference of latitude given, to find the distance, and departure from the meridian, (or difference of longitude.)

In Problem 2d, we have the course and distance given, to find the true (or meridional) difference of latitude, and the departure, (or difference of longitude.)

With respect to heights and distances: If we suppose in the first Problem before mentioned, the angle at A to be the angle which the top of any distant object makes with the surface of the earth, where we stand; the base to be the distance of the object, (on level ground,) and the perpendicular, the object's height; then we have the angle A, and the distance AB, to find the height BC; but this will serve only on level ground, and where the object is accessible.

The

The distance of any inaccessible object may be found by Problem 1st, of oblique Trigonometry: for, if we suppose the object at C, then, at two stations, as at A and B, take the bearing of the place; also, measure the stationary distance AB, and you will then have two angles and a side opposite to one of them, to find either of the ther sides.

To take the height of an Object standing on a hill, which is inaccessible.

22

120

122°

114°

F

At two stations, as at A and B, take the angles, viz. CAE and CBE, which the top of the object makes with an horizontal line, and that which the bottom of the object makes with the first station, at A, viz. DAE, then take DAE from CAE, and the remainder is CAD.

Note. When an angle is expressed by three letters, the middle one shews the angle. Now, suppose the stationary distance AB 120, the angle ACB 12o, and angle CBA 122°, then by Problem 1st of oblique Trigonometry, we have two angles and a side opposite to one of them given, to find the side AC. Therefore,

Note. I. subtracted 122° from 180°, and worked with the remainder, and in the following, 114° from 180°. Now, having found AC 489 5, suppose the angle CDA 114°, and the angle CAD 22°, and we have two angles and a side opposite to one of them, as before to find the perpendicular height of the object CD. Therefore,

As