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
A64224	Thesaurarium mathematicae, or, The treasury of mathematicks containing variety of usefull practices in arithmetick, geometry, trigonometry, astronomy, geography, navigation and surveying ... to which is annexed a table of 10000 logarithms, log-sines, and log-tangents / by John Taylor.	Taylor, John, mathematician.	1687 (1687)	Wing T534; ESTC R23734	190,995	514

to wit 64D 23M Comp. is 25D 37M is 9. 635833 Tang. is 9. 680768 so 64D 23M is the others Compl. whose Sine is 9. 955065 and his Tang. is 10 319231 so for any other PROP. III. To find the Secant of any Arch or Angle propounded In this little Book I have not room to set down the Tables of Artificial Secants at large as I have done with the Sines and Tangents Nevertheless I will not here omit to shew how they may be easily found out by the Tables of Sines The method is thus substract the Logarithm Sine of the Sines compl of an Angle from the double Radius of the Tables and the remainder shall be the Secant required As if I desire the Secant of 25D 37M I find the Logarithm-sine of his complement to be 9. 955065 which substracted from the double Radius that is 20. 000000 there remains 10 044935 which is the Secant of it and so the Secant of 64D 23M is 9. 955065 which is the Complement of the former because they both are Equal to 20. 000000 the double Radius and so may any other be found out CHAP. IV. Of GEOMETRY THE End and Scope of Geometry is to measure well for as Number or Multitude is the Subject of Arithmetick so is Magnitude that of Geometry to measure well is therefore to consider the Nature of every thing that is to be measured to compare such like things one with another and to understand their Reason and proportion and also their similitude And this is the End and Scope of Geometry I shall not trouble you with the Definitions of Geometry they being largely handled by many and herein every one meanly conversant in the study of the Mathematicks is acquainted but shall immediately fall in hand with the principal Propositions which chiefly concern the other following parts of this treatise SECT I. The Explication of some Geometrical Propositions PROP. I. To erect a perpendicular on any part of a line assigned LET the Line be A B and on the point D 't is required to raise a perpendicular to A B To operate which first open your Compasses to any convenient distance and placing one foot thereof in D with the other make the two marks C and E equidistant from D then open the Compasses to some other convenient distance and set one foot in E and describe the Arch FF then likewise in C describe the Arch GG then through the Intersections of these two Arches and to the point D draw H D perpendicular to A B as was required PROP. II. To Erect a Perpendicular on the End of a Line Let the given line be A B and on the End thereof at B 't is required to raise a Perpendicular line To perform which open your Compasses to the distance B D then on B as a Center describe the Arch D E F then from D to E place BD then placing one foot in E describe the Arch CF then remove your Compasses to F and draw the Arch CE Lastly through their Intersection draw C B which is a Perpendicular to AB on the end B as required PROP. III. From a Point above to let fall a Perpendicular on a Line Let the line given be B A and 't is required from the point above at C to let fall a Perpendicular to the said Line To perform which place one foot of your Compasses in C and open them beyond the given line A B and describe the Arch EF divide EF into two parts in D Lastly draw CD which shall be perpendicular unto AB falling from the point above at C as was so required PROP. IV. To draw a right line Parallel to a right line at any distance assigned Let the distance assigned be O E and the Lime given be A B and 't is required to draw C D Parallel to A B at the distance O E To perform which take in your Compasses the distance O E and on A describe the Arch H and on B the Arch K then draw C D so as it may justly touch the two Arches but cut them not so shall C D be parallel to A B at the assigned distance O E as was required PROP. V. To Protract an Angle of any Quantity of Degrees propounded Let it be required to Protract or lay down an Angle of 40 degrees To perform which first draw a right line as A B then open y●●r Compasses to 60 degrees in your line of Chords and with that Distance on A describe the Arch E F then take 40 degrees in your Compasses out of your line of Chords and place it on the Arch from F to H Lastly through the point H and from A draw A C so shall the Angle CAB contain 40 degrees as required PROP. VI. To measure an Angle already protracted Let the Angle given be C A B and 't is ●equired to find the Quantity thereof To ●erform which take in your Compasses 60 de●rees from your line of Chords and on A ●escribe the Arch EF then take in your Com●asses the Distance FH and apply it to your 〈…〉 ne of Chords and you will find the Angle 〈…〉 AB to contain 40 degrees PROP. VII To divide an Angle into two Equal parts Let the Angle given be BAC and 't is required to divide it into two equal parts To perform which do thus first take in your Compasses any convenient distance and placing one foot in A describe the Arch FKHE then on H describe the Arch KK and on K the Arch HH lastly through the Intersections of these two Arches draw the line AD to the Angular point A so shall the Angle BAC be divided into two equal parts viz. BA● and DAC as required PROP. VIII To divide a right line into any Number of Equal or Unequal parts or like to any divided line propounded Let the line A B be given to be divided into 5 equal parts as the line CD To perform which do thus first on the point C draw out a line making an Angle with CD at pleasure then make CF equal to AB and joyn their Extremities FD then draw Parallel lines to FD through all the 5 points of CD by the 4 prop. aforegoing which shall divide AB into 5 equal parts as required This way is to be observed when the line given to be divided is greater than the divided line propounded CASE II. But if AB be shorter than the given divided line CD take the line AB in your Compasses and on D strike the Arch F then draw the Tangent CF then take the nearest distance from the first division of CD to the Tangent-line CF which distance shall divide AB into 5 equal parts as the given divided line CD as required PROP. IX How to Protract or lay down any of the Regular Figures called Polygons To perform which divide 360 degrees the number of degrees in a Circle by the number of the

Poligon his sides as if it be a Pentagon by 5 if a Hexagon by 6 c. the Quotient is the Angle of the Center its Complement to 180D or a Semi circle is the Angle at the Figure half whereof is the Angle of the Triangle at the Figure Now I will shew how to delineate any Poligon three ways viz. 1 by the Angle at the Center 2. by the Angle at the Figure 3. by the Angle of the Triangle at the Figure I have hereunto annexed a Table which gives at the first sight without the trouble of Division 1. the quantity of the Angle at the Center 2. the quantity of the Angle at the Figure and 3 the Quantity of the Angle at the Triangle of the Figure from a Triangle to a Decigon Names of the Poligons Sides Angles at the Center Angles at the Figure Angles at the Trian D M D M D M Triangle 3 120 00 60 00 30 00 Square 4 90 00 90 00 45 00 Pentagon 5 72 00 108 00 54 00 Hexagon 6 60 00 120 00 60 00 Heptagon 7 51 43½ 128 34½ 64 17¼ Octogon 8 45 00 135 00 67 30 Nonigon 9 40 00 140 00 70 00 Decigon 10 36 00 144 00 72 00 CONSTRUCTION I. First by the Angle at the Center to delineate a Hexagon whose Angle at the Center is 60 degrees first lay down an Angle of 60 deg by prop. the 5. aforegoing making its sides of a convenient length at pleasure then take such a distance from O the Center of the figure equally on both sides as may make the third side equal to the side of the Poligon given which here is 100 parts Then divide the third side equally into two equal parts and draw a line through it from ☉ the Center set each half of the side of the Poligon 100 to wit 50 on each from the middle of the third line thus having placed the side of the Hexagon PP 100 parts in order describe the whole Hexagon PPPPPP as was required CONSTRUCTION II. Now by the Angle of the Figure to delineate any regular Poligon Let it be required to protract a Hexagon whose side as afore is 100 parts first I draw a line and make it 100 of those parts then I sind in the precedent Table the Angle of a Hexagon at the figure to be 120 degrees Then on each side of the drawn line I lay down an Angle of 120 deg according to the 5 precedent propositions and so work 6 times or as many times as your Poligon hath sides making each side 100 parts and each Angle 120 degrees so shall you have enclosed the Poligon PPPPPP as required CONSTRUCTION III. To Protract or lay down a Hexagon or any other regular Poligon by the Angle of the Triangle do thus First draw the side of the Hexagon P P make it 100 parts I find in the precedent Table that the Angle of the Triangle is 60 deg then at each end of the line P P I lay down an Angle of 60 deg by prop. 5. precedent and continue the two lines PO and PO untill they intersect each other in O then on O as a Center OP being Radius describe a Circle and within it describe the Hexagon PPPPPP as you see in the figure and so may you delineate any other Poligon whose Angels from a Triangle to a Decigon are all specified in the precedent Table PROP. X. To divide a line according to any assigned proportion Admit the right line given to be AB and 't is required to divide the same into two parts bearing proportion the one to the other as the lines E and F doth To perform which first draw the line CD equal to the given line AB Then draw the line HC from C to contain an Angle at pleasure Then from C to G place the line F and from G to H place the line E Then draw the line HD And lastly draw GK parallel to HD by the 4 prop. precedent so is the line DC equal to AB and divided into two parts bearing such proportion to each other as the two given lines E and F as was required PROP. XI To two lines given to find a third proportional to each of them Admit the two given lines be A and B and 't is required to find a third proportional to A as A to B First make an Angle at pleasure as HIK Then place the line B from I unto P and the line A from I unto L and draw PL. then also place the line A from I unto M and draw QM parallel unto LP by 4 prop. so shall the line IQ be a third proportional unto the two given lines A and B as was required For as B is to A so is A unto the proportional found IQ PROP. XII To three lines given to find out a fourth proportional unto them Admit the three given lines to be A B and C and 't is required to find a third proportional to them which shall have such proportion unto A as B hath unto C. To perform which first make an Angle at pleasure as DKG now seeing the line C hath such proportion to B as the line A unto the line sought Therefore place the line C from K unto H and B from K to F and draw FH Again place the line A from K to I and draw IE parallel unto FH by 4 prop. until it cutteth DK in E so have you the line KE a fourth proportional as was required For as C is unto B so is A unto the found line KE PROP. XIII To find a mean proportional Line between any two right lines given Let the two given lines be A and B between which it is required to find a mean proportional line To perform which first joyn the two lines A and B together so as they make the right line CED Then describe thereon a Semicircle CFD Then on the point E erect the perpendicular EF by 1 prop. to cut the limb of the Semi-circle in F so shall EF be a mean proportional line between the two given lines A and B as required PROP. XIV To find two mean proportional Lines between any two right Lines given Let the two given lines be A and B between which 't is required to find two mean proportionals To perform which first make an Angle containing 90 deg making the sides CD and CE of a convenient length then from C place the line B unto F and the line A from C unto G and draw FG which divide equally in H and describe the Semi-circle F K G. Then take the line B in your Compasses and placeing one soot in G with the other make a mark in the limb of the Semi-circle in K then draw ST in such sort that it may justly touch the Semi-circle in K and may cut through the two sides of the Angle equidistant from the Center of the Semi-circle H so shall SF and TG be two