II. The perpendicular radius, AB will be the tangent of c, or the co-tangent of A; and AC will be the secant of c, or the cosecant of A.

1. Extend the compasses from 384 to 288 on the line of numbers, that extent will reach from 45° to 53°. 8' on the line of tangents.

2. Extend the compasses from 53°. 8′ to 90° on the line of sines, that extent will reach from 384 to 480 on the line of numbers.

(S) Plane sailing in navigation is nothing more than the practice of rightangled trigonometry; calling the hypothenuse the distance sailed, the perpendicular the difference of latitude, the base the departure, and the angle opposite the base the course.

In the annexed figure, w ESN represents the horizon of the place c, from whence the ship sails, CA the point of

to.

the compass or rhumb she sails on, and the place sailed Then we represents the parallel of latitude sailed from, and AA the parallel of latitude arrived at. Hence BC be

comes

comes the difference of latitude, AB the departure, ca the distance sailed, the angle ACB the course, and the angle BAC the complement of the course. And this will always be the case whether the ship sails between the north and west, viz. between N and w; the north and east, viz. between N and E; the south and west, viz. between s and w; or south and east, viz. between s and E.

་

A

B

(T) The practice of Mercator's sailing is only the application of right-angled trigonometry, and similar triangles (Y.33 and 34). What is called the difference of latitude in plane sailing, is here called the proper difference of latitude, to distinguish it from the line CD, which is called the meridional difference of latitude. The part BD, by which CB is enlarged, is found by help of a table of meridional parts; being the sum, or difference of the meridional parts answering to the latitude sailed from and bound to; according as they are on different sides, or on the same side of the equator. ED is the difference of longitude.

E

CHAP. II.

INVESTIGATION OF RULES FOR CALCULATING THE SIDES AND ANGLES OF OBLIQUE ANGLED TRIANGLES, AND THE LOGARITHMICAL SOLUTIONS OF ALL THE CASES.

PROPOSITION II.

(U) The excess of the greater of two given magnitudes above half their sum, is equal to half the difference between those magnitudes ; or half their sum increased by half their difference, gives the greater magnitude; and being diminished by half their difference, leaves the less magnitude.

Let AC and CB be two unequal mag- £ nitudes, of which AC is the greater,

B

Bisect AB in D and make AE=CB, then will AD half the sum of AC and CB, and DC half the difference; for, since AB is bisected in D, and AE=CB, it follows that EC is bisected in D, and EC is the difference between AC and CB.

Hence AC-AD=DC; or AD+DC=AC, and AD-ED(=DC)= AE(BC). Q. E. D.

PROPOSITION III.

(W) In any plane triangle, the sine of any angle, is to the

side

side opposite to it, as the sine of any other angle, is to the side opposite to it, and the contrary.

Since any triangle whatever may be inscribed in a circle, let ABC be the given triangle and o the centre of the circle. Draw OD and OE per- A pendicular to the sides of the triangle, then (by Euclid III. and 3. and III. and 30.) the sides of the

E

triangle will be bisected, and also the arcs which these sides subtend. Now an angle at the centre of a circle is double to an angle at the circumference when they stand on the same arc, (Euclid III. and 20.) therefore an angle at the centre is equal to an angle at the circumference standing on the double arc; or in other words, an angle at the circumference of a circle is measured by half the arc on which it stands, and an angle at the centre by the whole arc on which it stands: hence the angle coD=the angle CAB, and COE=CBA, &c. Now cI is the sine of the angle cop and cm is the sine of the angle coE. Draw mi, which will be parallel to AB, then per similar triangles cr : CB :: cm: AC, viz. the sine of the angle at A is to the side BC, as the sine of the angle at B is to the side AC, and the contrary by inversion.

COROL. The sides of triangles are to each other as the chords of double their opposite angles.

PROPOSITION IV.

(X) 1. In any plane triangle, the sum of any two sides is to their difference; as the tangent of half the sum of their opposite angles, is to the tangent of half their difference.

Let ABC be any triangle; make BE=BC and join CE, then the angle CBE, being the exterior angle of the triangle ABC, is equal to CAB + BCA (Euclid I.

and 32.)

Bisect AE in D and CE in G, then join DG and it will

be parallel to AC; draw BF

parallel to DG, join BG, and with the centre B and radius BG describe a circle.

The angle CBG the angle EBG equal to half the sum of the angles CAB and BCA; for the triangles CBG and EBG have the two sides BC and CG, equal to the two sides BE and EG, and the side BG common to both, therefore they are equal in all respects (Euclid I. and 8.)

BF being parallel to AC, the angle FBE the angle CAB, and the angle CBF the angle ACB. (Euclid I. and 29.)

The angle GBF half the difference between the angles CBF and FBE, for it is the excess of the greater above half their sum

GBE

GBE (U. 43); hence GE=tangent of the half sum, and GF= tangent of the half difference between the two angles CBF and FBE. For the same reason, DB being the excess of the greater of the two given lines AB and BE or BC, above half their sum AD, it is equal to half their difference.

By similar triangles, DE (

AB+BC
2

): DB (

=3

(= tangent ACB + CAB): GF (=tangent

ACB-CAB) Q.E.D.

Half the sum of the angles ACB and CAB added to half their difference gives the angle opposite to the greater side (U. 43.), for the greater angle is opposite to the greater side: and half the sum diminished by half the difference, gives the angle opposite to the less side; for the less angle is opposite to the less side, (Euclid I. and 18.)

(Y) NOTE. Instead of the tangent of half the sum of the opposite angles, the tangent of half the supplement of the contained angle may be used; or the co-tangent of half the contained angle *, which are all equal to each other.

PROPOSITION V.

(Z) In any plane triangle, double the base or longest side, is to the sum of the other two sides, as their difference is to the distance of a perpendicular from the middle of the base: this distance, added to half the base, gives the greater segment, and being subtracted leaves

the less..

Let ABC be the triangle; with c as a centre, and the radius CB=the least side, describe a circle, produce ac to H; then because CF CB=CH; AH=AC+CB the

sum of the sides, and AFCBC the A G E D

difference between the sides.

H

B

Because CD is perpendicular to GB, GD=BD (Euclid III. and 3.) therefore AG is the difference between the segments of the base; but ED AG (U. 43.) for it is the excess of the greater segment AD above half their sum AE; AB being the base, or sum of the segments AD and DB, and AE its half.

Then (Euclid III. and 36 corol.) AHX AFABX AG, there fore AB: AH : AF: AG, that is AB AC + CB :: AC-CB: 2ED,

Since the sum of the three angles of every triangle is equal to 'two right angles, it is plain that half the sum of any two angles is half the supplement of the third angle. And, that the tangent of half the supplement of any angle is equal to the cotangent of half that angle may be thus shewn :

Let HBr (Fig. 1. Plate 1.) be any arc bisected in B, and FAƒ, its supplement, bisected in ; then AK, the cotangent of the half arc Br, is evidently the tangent of the half supplement AF.

or,

As the sine of any other angle,
Is to its opposite side.

(W. 43.)

(E) An angle found by this rule is sometimes ambiguous, for trigonometry gives us only the sine of an angle, and not the angle itself, and the sine of every angle is also the sine of its supplement. The tables give only the acute value of an angle; the obtuse value is the supplement thereof.

When the given side, opposite to the given angle, is greater than the other given side; then the angle opposite to that other given side is always acute. But when the given side opposite to the given angle, is less than the other given side, then the angle opposite that other given side may be either acute or obtuse, and consequently it is ambiguous.

(F) II. When two sides and the angle contained between them are given.*

The sum of the two given sides,

Is to their difference;

As the co-tangent of half the contained angle,

Is to the tangent of half the difference between the other two angles.

This half difference added to the complement of half the contained angle, gives the angle opposite to the greater side; and being subtracted, leaves the angle opposite to the less side.

OR,

The sum of the two given sides,

Is to their difference;

As the tangent of half the supplement of the contained angle,
Is to the tangent of half the difference between the other

two angles.

This half difference, added to the half supplement, gives the angle opposite to the greater side; and being subtracted, leaves the angle opposite to the less side. (X. 44.)

The remaining side of the triangle may be found by Rule I. (G) III. When the three sides of a triangle are given to find the angles.t

Double the base, or longest side,

Is to the sum of the other two sides;

As the difference between those two sides,

Is to the distance of a perpendicular from the middle of the base.

*If the two given sides be equal to each other, the triangle is isosceles, and each of the remaining angles will be equal to half the supplement of the given angle. If the given angle be 90°, the required parts may be found by. Case VI. of right angled triangles.

† If the triangle be equilateral each of its angles will be 60°; if it be isosceles the perpendicular will bisect the base, and the angles may be found by Case IV. of right angled triangles.

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