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f. 28 Dat. pofition cut one another is given f. and the ftraight line DB which g. 33. Dat. is at right angles to AB is given in pofition, and AB is given in pofition, therefore f the point B is given. and the points A, D are given, wherefore h the ftraight lines AB, BD are given. and the ratio of AB to BC is given, and therefore i BC is given.

h. 19. Dat. i. 2 Dat.

k. 12. 5.

d. 47. X.

The Compofition.

Let the given ratio of FG to GH be that which AB is required to have to BC, and let IK be the given ftraight line which is to be taken from BC, and let the 'ratio which the remainder, is re-. quired to have to BD be the given ratio of HG to GL, and place GL at right angles to FK, and join LF, LH. Next, as HG is to

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F G H

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GF, fo make HK to AE; produce AE to N fo that AN be the ftraight line to the fquare of which the fum of the fquares of AB, BD is required to be equal; and make the angle NED equal to the angle GFL. from the center A at the diftance AN defcribe a circle, and let its circumference meet ED in D, and draw DB perpendicular to AN, and DM making the angle BDM equal to the angle GLII. laftly, produce BM to C fo that MC be equal to JIK. then is AB the firit, BC the fecond and BD the third of the fight lines that were to be found.

For the triangles EBD, FGL, as alfo DBM, LGH being equiangular, as IB to BD, to is FG to GL; and as DD to BM, to is LG to CH; therefore, ex accuali, as LB to BM, fo is (FG to GH, and fo is) AE to IIK or MC; wherefore k AB is to EC, as AL to HK, that is, as FG to GH, that is, in the given ratio. and from the thaight line BC taking MC which is equal to the given ftraight line HK, the remainder DM has to BD the given ratio of HG to GL. and the fam of the fquares of AB, BD is equal the fquare of AD or AN which is the given space. Q. E. D. I believe it would be in vain to try to deduce the preceding Conftruction from an Algebraical Solution of the Problem.

FIN I S.

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THE

ELEMENT S

O F

PLANE AND SPHERICAL

TRIGONOMETRY.

GLASGOW :

PRINTED AND SOLD BY ANDREW FOULIS; SOLD ALSO BY

ROBERT CROSS, NEAR THE COLLEGE.

M.DCC.LXXXI.

PLANE TRIGONOMETRY;

LE

LEMMA I. FIG. I

as

ET ABC be a rectilineal angle, if about the point B center, and with any diftance BA, a circle be described, meeting BA, BC, the straight lines including the angle ABC in A, C; the angle ABC will be to four right angles, as the arch AC to the whole circumference.

Produce AB till it meet the circle again in F, and through B draw DE perpendicular to AB, meeting the circle in D, E.

By 33. 6. Elem. the angle ABC is to a right angle ABD, as the arch AC to the arch AD; and quadrupling the confequents, the angle ABC will be to four right angles, as the arch AC to four times the arch AD, or to the whole circumference.

LEMMA II. FIG. 2.

LET ABC be a plane rectilineal angle as before: about B as a center with any two diftances BD, BA, let two circles' be described meeting BA, BC in D, E, A, C; the arch AC wilf be to the whole circumference of which it is an arch, as the arch DE is to the whole circumference of which it is an arch.

By Lemma 1. the arch AC is to the whole circumference of which it is an arch, as the angle ABC is to four right angles and by the fame Lemma 1. the arch DE is to the whole cir cumference of which it is an arch, as the angle ABC is to four right angles; therefore the arch AC is to the whole circumfe rence of which it is an arch, as the arch DE to the whole cir cumference of which it is an arch.

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LET ABC be a plane rectilineal angle; if about B as a center, with BA any distance, a circle ACF be defcribed meeting BA, BC, in Á, C; the arch AC is called the measure of the angle ABC.

II.

The circumference of a circle is fuppofed to be divided into 360 equal parts called degrees, and each degree into 60 equal

, parts called minutes, and each minute into 60 equal parts called feconds, &c. And as many degrees, minutes, feconds, &c. as are contained in any arch, of fo many degrees, minutes, feconds, &c. is the angle, of which that arch is the measure, faid to be.

COR. Whatever be the radius of the circle of which the meafure of a given angle is an arch, that arch will contain the fame number of degrees, minutes, feconds, &c. as is manifeft from Lemma 2.

III.

Let AB be produced till it meet the circle again in F, the angle CBF, which, together with ABC, is equal to two right angles, is called the Supplement of the angle ABC.

IV.

A ftraight line CD drawn through C, one of the extremities of the arch AC, perpendicular upon the diameter paffing through the other extremity A, is called the Sine of the arch AC, or of the angle ABC, of which it is the meafure.

COR. The Sine of a quadrant, or of a right angle, is equal to the radius.

V.

The fegment DA of the diameter paffing through A, one extremity of the arch AC between the fine CD, and that extremity is called the Versed Sine of the arch AC, or angle ABC.

VI.

A ftraight line AE touching the circle at A, one extremity of the arch AC, and meeting the diameter BC paffing through the other extremity C in E, is called the Tangent of the arch AC, or of the angle ABC.

VII.

The ftraight line BE between the center and the extremity of the tangent AE, is called the Secant of the arch AC, or angle ABC.

COR. to def. 4. 6. 7. The fine, tangent, and fecant of any angle ABC, are likewife the fine, tangent, and fecant of its fupplement CBF.

It is manifeft from def. 4. that CD is the fine of the angle CBF. Let CB be produced till it meet the circle again in G; and it is manifeft that AE is the tangent, and BE the fecant, of the angle ABG or EBF, from def. 6. 7.

Fig. 4. COR. to def. 4. 5. 6. 7. The fine, verfed fine, tangent, and

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