SECTOR. 



tance of 90 and 90 in tlie fines equal to the given radius ; 

 take the tranfverfe diflance of the fine complement of the 

 given degrees ; if the given degrees are lefs than 90, the dif- 

 ference, but if greater, the fum of the fine complement and 

 radius gives the verfed line. 



7. To open tile legs of the feftor fo that the correfponding 

 double fcales of lines, chords, fines, tangents, may make, 

 each of them, a right angle. On the lities, make the lateral 

 diftance 10, a diftance between 8 on one leg, and 6 on the 

 other leg ; on the fines, make the lateral dillance 90 a tranf- 

 verfe dillance from 45 to 45, or from 40 to 50, or from 30 

 to 60, or from the fine of any degrees to their complement ; 

 or, on the fines, make the lateral diftance of 45 a tranfverfe 

 diftance between 30 and 30. 



Sector, in Trigonometry, Ufc of the. I. The bafe and 

 perpendicular of a right-angled triangle being given, to find 

 the hypothenufe. Suppofe the bafe AC [Plate Trigo- 

 nometry, Jig. 5.) 40 miles, and the perpendicular A B 30 ; 

 open the fedlor till the two fcales of lines make a right angle ; 

 then, ior the bafe, take 40 parts on the fcale of lines on one 

 leg ; and, for the perpendicular, take 30 on the fame fcale on 

 the other leg ; then the extent from 40 on the one to 30 on 

 the other, taken in the compafles, will be the length of the 

 hypothenufe ; which line, applied to the fcale of lines, will 

 be found ^o miles. 



2. The perpendicular A B of a right-angled triangle 

 ABC being given, 30, and the angle B C A 37 degrees ; 

 to find the hypothenufe B C. Take the given fide A B, 

 and fet it over, on each fide, on the fine of the given angle 

 A C B ; then the parallel diftance of 90 and 90 or radius, 

 will be the hypothenufe B C ; which will mealure 50 on the 

 fcale of lines. 



3. The hypothenufe and bafe being given, to find the per- 

 pendicular. Open the feftor till the two fcales of lines be 

 at right angles ; then lay off the given bafe on one of thofe 

 fcales from the centre ; take the hypothenufe in your com- 

 pafTes, and fetting one foot in the term of the given bafe, let 

 the other fall on the fcale of lines on the other leg ; the dif- 

 tance from the centre to the point where the compafles fall 

 will be the length of the perpendicular. 



4. The hypothenufe being given, and the angle A C B ; 

 to find the perpendicular. Make the given hypothenufe a 

 parallel radius, i. e. make it the extent from 90 to 90 on tiie 

 fcales of fines ; then will the parallel fine of the angle A C B 

 be the length of the fide A B. 



J. The bafe and perpendicular A B given, to find the 

 angle B C A. Lay off the bafe A C, on both fides the 

 fe£tor, from the centre, and note its extent ; then take the 

 given perpendicular, and to it open the feftor in the terms of 

 the bafe ; the parallel radius will be the tangent of B C A. 



6. In any right-lined triangle, two fides being given, witli 

 the included angle ; to find the third fide. Suppofe the fide 

 A C (Jig. 6. ) 20, the fide B C 30, and the included angle 

 ACB I 10 degrees; open the feftor till the two fciles of lines 

 make an angle equal to the given angle, viz. 110 degrees; 

 lay off the given fides of the triangle, from the centre of the 

 feftor, on each of the fcales of lines ; the extent between tiieir 

 extremes is the leni;th of the fide A B fought. 



7. The ai'gles CAB and ACB given, and the fide C B ; 

 to find the bafe A B. Take the given fide C B, and turn 

 it into the parallel fme of its opjiofi'.e angle CAB; and tiien 

 the parallel fine of the ancle ACB will be the length of the 

 bafe A B. 



8. The three angles of a triangle being given ; to find the 

 proportion of the fides. T:ike the lateral fines of the 

 feveral angle?, and meafure tiiem in the fciile of lines ; the 

 numbers anfwcring to which give the proportion of the fides. 



9. The three fides being given, to find the angle ACB. 

 Lay the fides AC, C B, along the fcales of lines, from 

 the centre, and fet over the fide A B in their terms ; fo i. 

 the feftor opened, in thefe lines, to the quantity of the 

 angle ACB. 1 J 



10. The hypothenufe AC {fg. 7.) of a right-angled 

 fphencal triangle ABC, given, r. g. 43 degrees, and the 

 angle CAB 20 degrees ; to find the fide C B. The rule 

 is, as radius is to the fine of the given hypothenufe 43 de- 

 grees, fo ii the fine of the given angle 20 degrees to the fine 

 of the perpendicular C B. Take then 20 degrees from the 

 centre, along the fcale of fines, in your compaffcs, and fet 

 the extent from 90 to 90 on the two legs ; and the parallel 

 fine of 43 degrees, the given hypothenufe, will, when mea- 

 fured from the centre on the fcale of fines, give 13° 30', the 

 fide required. 



11. The perpendicular B C, and the hypothenufe AC, 

 given, to find the bafe A B. As the fine complement of 

 the perpendicular B C is to radius, fo is the fine comple- 

 ment of the hypothenufe to the fine complement of the 

 bafe. Therefore make the radius a parallel fine of the 

 complement of the given perpendicular, e.gr. 76° 30' ; then 

 the parallel fine of the complement of the hypothenufe, e. gr. 

 47°, meafured along the fcale of fines, will be found 49' 

 25', the complement of the bafe required ; confequently the 

 bafe itfelf will be 40'' 35'. 



Sector, in Geometry, y<r. particular u/et of the. I. To 

 make a regular polygon, whole area (liall be of any given 

 magnitude. Let the fii^'ure required be a pentagon, whofo 

 fuperficial area is 125 feet ; extraft the fquare root of .^ of 

 125, it will be found 5. Make a fquare, whofe fide is 

 5 feet ; and, by the line of polygons, as already direfted, 

 make the ifofceles triangle C G 1) [Plate Yl.W\. Geometry, 

 Jig. 8.) fo as that C G being the femi-diameter of a circle, 

 C D may be the fide of a regular pentagon infcribed in it ; 

 then let fall the perpendicular G E. Then continuing the 

 lines E G and E C, make E F equal to the fide of the fquare 

 before made ; and from the point F, draw the right hne 

 F H parallel to G C ; then a mean proportional between 

 G E and E F will be equal to half the fide of the polygon 

 fought, which, doubled, will give the whole fide. The 

 fide of the pentagon thus had, the pentagon itfelf may be 

 defcribed, as above direfted. 



2. A circle being given, to find a fquare equal to it. 

 Divide the diameter into fourteen equal parts, by the fcale 

 of lines, as above direfted; then will 12.4 of tiiofe parts, 

 found by the fame line, be the fide of the fquare fought. 



3. A fquare being given, to find the diameter of a circle 

 ecHial to it. Divide tlie fide of the Iquare into eleven equal 

 parts, by means of the fcale of lines ; and continue that 

 fide to 1 2.4 parts; this will be the diameter of the circle re- 

 quired. 



4. To find the fide of a fquare equal to an ellipfis, whofr 

 tranfverfe and conjugate diameters ;irc given. Find a mean 

 proportional between the tranlverfe and conjugate diameters; 

 which being divided into fourteen equal parts, I2,*j of it 

 will be the fide of the fquare required. 



5. To dcfcribe an ellipfis in any given ratio of its dia- 

 meter, the area of which fliall be equal to a given (qiiare. 

 Suppofe the proportion of the tranlverfe and conjugate dia- 

 meters be required, as 2 to i ; divide the lide of the given 

 fquare into eleven equal parts ; then, as 2 is to I, fo is 1 1 

 X 14 =: 154 to a fourth numl)er ; the fquare root of which 

 is the conjugate diameter fought. Then, .is 1 is to 2, fo is 

 the conjugate diameter to the trurfvrrfc. Now, 



6. To dclcrlbe an ellipfis, by having the tranlverfe and 

 conjugate diameters given. Let the two diameters A D, 



