and C, and it is done; for the angle C will be 320 22′, the angle A 57° 38', and BC 143,6.

Extend from the hypotenuse 170 to the perpendicular 91 on the line of numbers; that extent will reach from radius to the angle C, or the complement of angle A=32° 22′ on the line of sines.

2dly. Extend from radius to the angle A 57° 38' on the line of sines; that extent will reach from the hypotenuse 170 to the base 143.6, on the line of numbers.

CASE VI.

The legs given, to find the angles and hypotenuse.

The legs AB 890, BC 707, given, to find the angle BAC or ACB, and the hypotenuse AC.

BY PROJECTION.

Make BC=707, and on B erect the perpendicular BA, which make equal to 890; join AC, and it is done; for the angle C will be 51° 32'; consequently the angle A 38° 28', and the hypotenuse 1137.

068

B

The extent from 707 to 890 on the line of numbers will reach from radius, or 45 degrees, to the angle C 51° 32′ on the line of tangents.

2dly. The extent from the angle C 51° 32′ to radius, or 90°, on the line of sines, will reach from the base 890 to the hypotenuse 1137, on the line of numbers.

* When you take the log. sines, or tangents to the nearest minute only, it is best to use this canon for finding BC, which is more correct than the one found by making the perpendicular radius; because the variation of the log. sine of an arch is less than the corresponding variation of the log. tangent When you are finding AC it is best to make the greatest side radius, for the reason mentioned in the last note.

QUESTIONS

To exercise the learner in Right-Angled Plane Triganometry...

Quest. 1. The hypotenuse 496 miles, and the angle opposite to the base 560 15', given, to find the base and perpendicular.

Ans. Base 412.4, and the perpendicular 275.6 miles.

Quest. 2. The perpendicular 275 leagues, and the angle opposite to the base 56° 15′, given, to find the hypotenuse and base.

Ans. The hypotenuse 495, and base 411.6 leagues.

Quest. 3. The base 33 yards, and the angle opposite to the perpendicular 530 26', given, to find the hypotenuse and perpendicular.

Ans. Hypotenuse 55.39, and the perpendicular 44.49 yards.

Quest. 4. The hypotenuse 575, and perpendicular 50 miles, given, to find the base.

Ans. Base 572.8 miles.

Quest. 5. The hypotenuse 59, and the base 33 miles, given, to find the perpendicular.

Ans. Perpendicular 48.9 miles.

Quest. 6. The base 33, and perpendicular 52 leagues given, to find the hypotenuse.

Ans. Hypotenuse 1.59 leagues.

OBLIQUE TRIGONOMETRY.

CASE I.

Two angles and one side given, to find either of the legs.

Given the angle BAC=100°

the angle ACB 54° and the leg AB 220 to find the sides.

H

BY PROJECTION.

Subtract the sum of the angles A and C from 1800, the remainder will be the angle B=260. Draw the indefinite line BE, also the line BH, making the angle EBH=26°, on BH set off BA 220. On A make the angle BAC 100°; then AC will intersect the line BE in the point C, which completes the triangle, and BC will measure (on the same scale from which BA was laid down) 268 nearly, and AC 119.

BY LOGARITHMS by Theorem II.

The extent from the angle C=54° to the angle A or its supplement 80° on the sines, will reach from AB=220 to BC=268 on the line of numbers.

2dly. The extent from the angle C=54° to the angle B=269, on the sines, will reach from AB=220 to AC 119 on the line of numbers.

CASES II. AND III.

Tico sides and an angle opposite to one of them being given, to find the ather angles and the third side.

NOTE. It may be proper to observe, that if the given angle be obtuse, the angle sought will be acute: but when the given angle is acute

and opposite to a

lesser given side,

G

then it is doubtful

whether the required angle is acute or obtuse; it ought therefore to be given by the conditions of the problem.

EXAMPLE. Let there be given the side BC 410, the side AB 640, and the angle A 230, to find the other side AC, and the angles ABC, BCA.

BY PROJECTION.

Draw the indefinite line FE, make the angle DAE=254°, on AD set off AB=640, then on B, with 410 in your compasses taken from the same scale, describe an arch cutting FE in the points C and G, join BC, BG, and it is done; for the triangle may be either ACB, or AGB, according as the angle C, or G, is acute or obtuse; if that angle be acute, the triangle will be ABC; the side AC will measure 908, the angle ACB will measure 3840, and the angle ABC will measure 1180 nearly; but if the angle at the base be obtuse, the triangle will be AGB; the side AG will measure 266, the angle AGB will measure 14140 and the angle ABG 15°.

If the side BC had been given greater than AB, there could have been only one answer to this problem; for in that case, the point G would have fallen on the continuation of the line CA towards F, in which case the angle A of the triangle would become equal to FAB, instead of being equal to its supplement, as is required by the conditions of the problem.

BY LOGARITHMS, by Theorem II.

To find the angle C or G.

As the side BC 410

To find AC. 2.61278 As sine angle C 38 30′ 9.60070 Is to AB 640

9.79410

2.80618

2.80618 So is sine angle ABC 1182

9.94593

23 30

To sine ang. C 38 30 or G 141 30 9.79410 To the side AC 907,8
Angle A add 23 30

2.95801

BY GUNTER.

1st. The extent from BC=410 to AB=640, on the line of numbers, will reach from A=2340, to 5830, on the line of sines, which is equal to the angleC; its supplement 141° 30' being equal to the angle G.

2dly. The extent from the angle C=38° 30′ to 62° 0′ (the supplement of the angle ABC, 118° 0′) on the sines, will reach from AB=640 to 908, on the line of numbers; therefore the side AC=908.

Or, the extent from 38° 30' (the supplement of the angle G) to the angle ABG=15° 0' on the sines, will reach from AB=640 to 266, on the line of numbers: hence AG=266.

II.

CASES IV. AND V.

Two sides and their contained angle being given, to find either of the other an

gles and the third side.

Given the side AB and angle BAC 96° 0' BCA and CBA.

110

m. AC 80 m. E

to

find the angles

BY GUNTER.

1st. The extent from the sum of the sides 190 to their difference 30 on the line of numbers, will reach from the half sum of the angles B and C 420 to their half difference 80 5' on the line of tangents. The sum of which half sum and half difference gives the angle C 500 5′ and their difference the angle B 33° 55'; the greatest angle being opposite to the greatest side.

2dly. The extent from the angle B 33° 55', to the angle A 960 (or its supplement 840) on the line of sines, will reach from the side AC 80 to the side BC 142,6 on the line of numbers.

CASE VI.

The three sides of a plane triangle given, to find the angles.

The sides AB 85, BC 57, AC 108 given, to find the angles ABC, BAC, BCA. BY PROJECTION.

Draw the line AC, and make it equal to 108; take 85 in your compasses, and with one foot on the point A, describe an arch; then take the distance 57 in your compasses, and with one foot on C, describe another arch intersecting. the former arch in the point B; join AB, CB, and it is done. For the angle A being measured will be found 3140, B=97°, and the angle C=514° nearly.

BY LOGARITHMS, by Theorem IV.

Suppose BD to be drawn perpendicular to AC, and that AG=GC.

Having divided the triangle into two right-angled triangles, the hypotenuses and bases of which are given, we may find the angles by Theorem I.

1st. The extent from the base AC=108, to the sum of the sides 142, on the line of numbers, will reach from the difference of the sides 28, to twice DG 36.8 on the same line of numbers.

2dly. The extent from the hypotenuse AB=85, to the greater segment AD 72.4, on the line of numbers, will reach on the sines from the radius 90° to 58° 24' which is the complement of the angle BAD.

3dly. The extent from the hypotenuse CB 57, to the lesser segment DC 35.6 on the line of numbers, will reach on the sines from the radius 90° to 38° 39', which is the complement of the angle BCA.

This case may be solved without dividing the triangle into two right-angled triangles by Theorem V.