Selected quad for the lemma: end_n

Word A	Word B	Word C	Word D	Occurrence	Frequency Band	MI	MI Band	Prominent
end_n	draw_v	line_n	perpendicular_a	3,095	5	14.0786	5	true

Text snippets containing the quad

ID	Title	Author	Corrected Date of Publication (TCP Date of Publication)	STC	Words	Pages
A48344	Nine geometricall exercises, for young sea-men and others that are studious in mathematicall practices: containing IX particular treatises, whose contents follow in the next pages. All which exercises are geometrically performed, by a line of chords and equal parts, by waies not usually known or practised. Unto which the analogies or proportions are added, whereby they may be applied to the chiliads of logarithms, and canons of artificiall sines and tangents. By William Leybourn, philomath.	Leybourn, William, 1626-1716.	1669 (1669)	Wing L1925; ESTC R213612	96,458	212

length of the Perpendicular C A. The Analogie or Proportion is As the Sine of the Angle at C is to the Log. of A B So is the Sine of the Angle B to the Logar of C A. Or As the Radius is to the Logar of A B So is the Tangent of B to the Logar of C A. CASE IV. The Hypotenuse C B 225 the Angle C 53 degr 7 min. and the Angle at B 36 degr 53 min. given to finde the Base B A and the Perpendicular C A. DRaw a right Line C B containing 225 of your Line of equal parts then taking 60 deg out of your Line of Chords set one foot of the Compasses in B and with the other describe the Arch e d also the Compasses continuing at the same distance place one foot in C and with the other describe the Arch g f. Then from the Point B and through the Point d draw a right Line also from the Point C and through the Point f draw another right Line these two Lines will intersect or cross each other in the Point A forming the Triangle C A B. Lastly take the Line A B in your Compasses and applying it to your Scale of equal parts you shall finde it to contain 180 and that is the length of the Base A B. Likewise A C being taken in the Compasses and measured upon the Line of equal parts will be found to contain 135 which is the length of the Perpendicular C A. The Analogie or Proportion is As the Radius is to the Logarithm of C B So is the Sine of C to the Logarithm of A B And the Sine of B to the Logarithm of C A. CASE V. The Hypotenuse C B 225 and the Base A B 180 being given to finde the Perpendicular C A. DRaw a right Line A B containing 180 of your Scale of equal parts and upon the end A erect a Perpendicular A C. Then take out of your Scale of equal parts 225 the length of your Hypotenuse given and setting one foot of the Compasses in B with the other describe the Arch h k cutting the Perpendicular A C in C then draw the Line C B so have you constituted the Triangle C A B. Lastly take in your Compasses the length of the Line A C and apply it to your Line of equal parts where you shall finde that it will contain 135 and that is the length of the Perpendicular C A. The Analogie or Proportion is 1. Operation As the Logarithm of C B is to the Radius So is the Logarithm of A B to the Sine of C. 2. Operation As the Radius is to the Logarithm of C B So is the Sine of B the Complement of C to the Log. of C A. CASE VI. The Base A B 180 the Angle C 53 degr 7 min. and the Angle B 36 degr 53 min. being given to finde the Hypotenuse C B. DRaw a right Line A B containing 180 parts of your Scale for the Base of your Triangle and on the end A erect a Perpendicular A C. Then take 60 degr out of your Line of Chords and upon the Point B with that distance describe the Arch d e and because the Angle at B is 36 degr 53 min. take 36 degr 53 min. from your Line of Chords and set it upon the Arch from d to e. Then from B through the Point e draw a right Line till it meet with the Perpendicular before drawn which it will do in the Point C. And thus have you protracted your Triangle C A B. Lastly take in your Compasses the length of the Hypotenuse C B and measure it upon your Scale of equal parts and you shall finde it to contain 225. The Analogie or Proportion is As the Sine of C is to the Logarithm of A B So is the Radius to the Logarithm of C B. CASE VII The Base A B 180 and the Perpendicular C A 135 being given to finde the Hypotenuse C A. DRaw a right Line A B containing 180 equal parts and upon the end A erect the Perpendicular A C and out of your Scale of equal parts take the length thereof 135 which set from A to C and draw the Line C B which constitutes the Triangle C A B. Lastly take the length of the Hypotenuse C B in your Compasses and measuring it upon your Line of equal parts you shall finde it to contain 225. The Analogie or Proportion is 1. Operation As the Logarithm of A B is to the Logarithm of C A So is the Radius to the Tangent of B. 2. Operation As the Sine of B is to the Logarithm of C A So is the Radius to the Logarithm of C B. These are the severall Varieties or Cases that can at any time fall out in the Solution of Right-angled plain Triangles wherefore we will now proceed to the Solution of Oblique plain Triangles II. Of Oblique-angled plain Triangles THE Triangle which I shall make use of in the Solution of the severall Cases appertaining to an Oblique-angled plain Triangle shall be this following C D B in which     parts   D B the Base contains 335   C B the longer Side 271   D C the shorter Side 100   And   deg m. C the obtuse Angle contain 122 00 D the 2 acute Angles 43 20 B the 2 acute Angles 14 40 CASE I. Two Sides as the Base D B 335 and the Side C B 271 and the Angle D 43 degr 20 min. opposite to C B to finde the Angle at C opposite to the Base D B. DRaw a right Line D B representing the Base of your Triangle which by help of your Scale of equal parts make to contain 335. Then upon the Point D with the distance of 60 degr of your Line of Chords describe the Arch k l and from your Chords take 43 degr 20 min. the quantity of the Angle at D and set it upon the Arch-line from l to k drawing the Line C D. And because your other given Side B C contains 271 parts take 271 out of your Line of equal parts and setting one foot in B with the other describe the Arch m n crossing the former Arch k l in the Point C then draw the Line C B. So shall you have constituted the Triangle C D B. Lastly because it is the Angle at C that is required take 60 degr of your Chords and upon C describe the Arch g h and taking the distance between g and h apply it to your Line of Chords and you shall finde it to reach from the beginning thereof beyond the end of the Line wherefore take 90 degr the whole Line and set that distance from g to o then take the remainder of the Arch o h and measure that upon your Chord and you shall finde it to contain 32 degr which added to 90 degr make 122 degr and that is the quantity of the Angle at C which was required The Analogie or Proportion is As

and K H both parallel to the Line B A. Then taking in your Compasses the distance A 6 set one foot of them in C and with the other describe the Arch Z and draw the Line A X so that it onely touch the Arch Z. Then taking in your Compasses the distance A H set one foot upon the Line A C moving it along till the other foot being turned about will onely touch the Line A Z and where the point of the Compasses resteth upon the Line A C which it will doe at e through that Point draw the Line o e parallel to B A. So shall B o being measured upon your Line of Chords give you 69 degr 22 min. the quantity of the enquired Angle at B. CASE VIII The Base A C 27 degr 54 min. and the Hypotenuse A B 30 degr being given to finde the Perpendicular B C. The Analogie or Proportion is As the Co-sine of the Base A C 62 degr 6 min. is to the Radius So is the Co-sine of the Hypotenuse 60 degr to the Co-sine of the Perpendicular B C. HAving drawn your Quadrant A B C take out of your Line of Chords 62 degr 6 min. the Co-sine of the Base A C and set them from B to S. Also take from the Chords 60 degr the Co-sine of the Hypotenuse and set them from B to m and draw the Lines S T and m n both parallel to B A. Then taking the distance A T in your Compasses set one foot in C and with the other describe the Arch V and draw the Line A W so that it may onely touch the Arch V. Then taking A n in your Compasses move one foot thereof gently along the Line C A till the other being turned about doth onely touch the Line A W and where the Point resteth upon the Line C A which you will finde to be at c there make a mark and draw the Line c a parallel to B A. Lastly take the distance from B to a and measure it upon your Line of Chords where you shall finde it to contain 78 degr 30 min. the Complement of the Perpendicular or C A measured upon the Chord will give you 11 d. 30 min. the Perpendicular it self These Five last are all the Cases in a Right-angled Sphericall Triangle that are resolvable by Sines alone Those which follow are to be resolved by Sines and Tangents joyntly and so will require another manner of Operation then the former CASE IX The Hypotenuse A B 30 degr and the Angle at the Base A 23 d. 30 min. being given to finde the Angle at the Perpendicular B. The Analogie or Proportion is As the Radius is to the Co-sine of the Hypotenuse A B 60 deg So is the Tangent of the Angle at the Base A 23 degr 30 min. to the Co-tangent of the Angle at the Perpendicular B. FIrst draw a right Line as C H and upon one end thereof as at C erect the Perpendicular C A and with the distance of 60 degr of your Line of Chords upon the Centre C describe the Quadrant C A D. Also upon the Point A with 60 degr of your Chord describe the Quadrant A B C. Being thus prepared First take 60 deg the Co-sine of the Hypotenuse A B and set them from A to O also take 23 d. 30 m. the quantity of the Angle at A and set them from C to r and draw the Line A M F and the Line O G parallel to A C. Then take in your Compasses the distance between C and G and setting one foot in D with the other describe the Arch K and draw the Line A L so that it onely touch the Arch K. Then placing one foot of the Compasses in F take the least distance you can to the Line C L which set from C to E. Lastly draw the Line A E cutting the Quadrant C B in N. So the distance C N measured upon your Chords shall give you 20 degr 38 min. the Complement of the Angle at B which was required or the distance B N will give you 69 degr 22 min. the Angle it self CASE X. The Hypotenuse A B 30 degr and the Angle at the Base A 23 d. 30 min. being given to finde the Base A C. The Analogie or Proportion is As the Radius is to the Co-sine of the Angle at A 66 d. 30 m. So is the Tangent of A B the Hypotenuse 30 degr to the Tangent of the Base A C. HAving prepared your Quadrants out of your Line of Chords take 66 degr 30 min. the Complement of the Angle at A and set them from A to a and draw the Line a b parallel to A C also take 30 degr the Hypotenuse from your Chord and set them from C to c and draw the Line A c prolonging it to d. This done take in your Compasses the distance from C to d and setting one foot in D with the other describe the Arch P and draw the Line C Q so that it may onely touch the Arch P. Then setting one foot of the Compasses in b take the least distance between b and the Line C Q which distance will reach from C to e. Lastly draw the Line A e cutting the Quadrant A C B in the Point M. So shall the distance C M being taken in the Compasses and measured upon the Line of Chords contain 27 degr 54 m. which is the quantity of the Base required CASE XI The Base A C 27 degr 54 min. and the Angle at the Base A 23 degr 30 min. being given to finde the Perpendicular B C. The Analogie or Proportion is As the Radius is to the Sine of the Base A C 27 degr 54 min. So is the Tangent of the Angle at A 23 degr 30 min. to the Tangent of the Perpendicular B C. YOur Quadrants being prepared out of your Line of Chords take 27 d. 54 m. the quantity of the given Base A C and set them from A to S drawing the Line S R parallel to C D. Also take out of your Chords 23 degr 30 minutes the quantity of the given Angle at A and set them from C to r and draw the Line A r prolonging it to F. This done take in your Compasses the distance A R and setting one foot in D with the other describe the Arch T and by the side thereof draw the Line C V onely to touch it Then set one foot of your Compasses in F and with the other take the nearest distance to the Line C V which distance will reach from C to X. Lastly draw the Line A X which will cut the Quadrant A B C in the Point Z. So shall C Z being measured upon the Line of Chords contain 11 degr 30 min. which is the quantity of the Perpendicular B C which was required CASE XII The Base A C 27 degr 54 min. and the Perpendicular B C 11 d. 30 min. being given to finde the Angle at the Base A. The

Point X. So a Line drawn from Q to X shall be perpendicular to the given Line R S. PROP. IV. A right Line being given to draw another right Line which shall be parallel thereto at any distance required LET A B be a Line given and let it be required to draw another right Line which shall be parallel thereunto and at the distance of the length of the Line C. First take the length of the Line C in your Compasses and setting one foot towards one end of the given Line as at D describe the Arch E. Secondly set one foot of the Compasses towards the other end of the given Line as at F and describe the Arch G. Lastly lay a Ruler to the Arches E and G so that the Ruler onely touch the Arches and not cut or cross them in any part So a Line drawn thereby shall be parallel to the given Line A B and at the distance of the Line C. PROP. V. A right Line being given to draw another right Line parallel thereunto which shall pass through a given Point LET the given Line be H K to which let it be required to draw a parallel Line which shall pass through the Point L. First take in your Compasses the distance K L and setting one foot of the Compasses in H with the other describe the Arch M M. Secondly take the Line H K in your Compasses and setting one foot in the given Point L with the other cross the Arch M M in the Point O. So a Line drawn from L to O shall be parallel to the given Line H K and shall pass through the Point L. PROP. VI. Three right Lines being given to make a Triangle whose three Sides shall be equal to the three given Lines LET the three Lines given be N P Q. First take the Line N in your Compasses and lay that down from R to S. Secondly take the Line P in your Compasses and setting one foot in S with the other describe the Arch V V. Thirdly take the Line Q in your Compasses and setting one foot in R with the other cross the Arch V V in the Point T. Lastly draw the Lines T R and T S. So shall you have constituted the Triangle T R S whose three Sides are equal to the three given Lines N P Q. PROP. VII Three Points which lie not in a straight Line being given to finde the Centre of a Circle which being described shall pass through the three given Points LET the three given Points be A B C. First open your Compasses to any distance greater then half the distance between A and B and setting one foot in B with the other describe the Arch G D. Then remove the Compasses set one foot in A and with the other cross the former Arch in the Points D and F and draw the Line D F. Secondly the Compasses still continuing at the same distance set one foot in the Point C and with the other cross the Arch before drawn in the Points E and G and draw the Line E G crossing the other Line D F in the Point H. So shall H be the Centre of the Circle which being described shall pass directly through the three given Points A B C. PROP. VIII Two Points within any Circle being given how to describe the Arch of another great Circle which shall pass through those two given Points and also divide the Circumference of the given Circle into two equal parts LET the two Points given be E and F within the Circle A B C D. First through either of them as through E draw the right Line E D passing through the Centre of the Circle at K. Secondly draw the Line A C at right Angles to B D so shall the Circle be divided into four equal parts or Quadrants by the Lines A C and B D. Thirdly draw the Line E A and upon the Point A by the II. Prop. erect the Perpendicular A G cutting the Line B D it being extended in the Point G so have you three Points E F and G through which by the last Prop. ou may draw a Circle to pass whose Centre will be at H upon which Point if you describe an Arch of a Circle at the distance H E or H F it will pass through the two given Points E and F and divide the Circle A B C D into two equal parts in the Points L and M which was required And that this Arch thus drawn doth divide the Circle into two equal parts is evident for a Line drawn from L to M will pass directly through the Centre K. These are such Geometricall Propositions as are absolutely necessary for the working of the severall Conclusions in the following Exercises More might have been added but these well understood and practised will be sufficient to carry you through this Work And now I will describe unto you the making of the slight Instrument by which all contained in this Book is performed namely The Line of Chords and shew you the general Vse thereof in the protracting or laying down of Angles of any quantity Or if any Angle be already laid down to finde thereby the quantity thereof Then will I give you some few usefull and necessary Theorems chiefly appertaining to Trigonometry or the Solution of Triangles and so conclude this first EXERCISE How to make a Line of Chords ACcording to the largeness of your Line of Chords you intend to make draw a right Line as A B and upon the Point A by the II. Prop. erect the Perpendicular A C and upon A as a Centre describe the Quadrant B D E C which you must divide into 90 equal parts or Degrees Which that you may readily doe your Compasses being opened to the distance A B set one foot in B and the other will reach to E also set one foot in C and the other will reach to D so is your Quadrant divided into three equal parts each part containing 30 degr This done divide each of these three parts into three more so shall you have divided your Quadrant into 9 equal parts each containing 10 degr and each of these 9 parts being divided into halves will contain 5 degr and if you make your Line large enough you must divide those into 5 equal parts which you may very well doe if the Line A B be but two inches long as all the Schemes and Figures in these Exercises are drawn by a Line of Chords of that length Your Quadrant being thus divided into 90 degr draw the Line B C and parallel thereto two other Lines one pretty close to B C to contain the small Divisions and the other at a larger distance to set the Figures in Now it is the Line B C which is called the Line of Chords possibly for this Reason the Arch or Ark B D E C representing Arcus a Bow and B C the String or Chord thereof the divisions whereof are to be transferred from the degrees of the Quadrant