An introduction to plain trigonometry, with its application to heights and distancesM. Heavisides, for the author, 1792 - 99 Seiten |
Im Buch
Ergebnisse 1-5 von 39
Seite 5
... Draw a ftraight Line through C and D , the Points of Interfection , which will cut the Line A B , into two equal Parts PROBLEM . 2 . To raife a perpendicular , at or near the Middle of a given right Line A B. 1. From the given Point g ...
... Draw a ftraight Line through C and D , the Points of Interfection , which will cut the Line A B , into two equal Parts PROBLEM . 2 . To raife a perpendicular , at or near the Middle of a given right Line A B. 1. From the given Point g ...
Seite 5
... draw a perpendicular , at the end of a given right Line . 1. With one Foot in D open the Compaffes to any Radius ... draw the Diameter C F. 4. Draw F D , which will be the perpen- dicular required . Problem 4 . PROBLEM 4 . * How to make ...
... draw a perpendicular , at the end of a given right Line . 1. With one Foot in D open the Compaffes to any Radius ... draw the Diameter C F. 4. Draw F D , which will be the perpen- dicular required . Problem 4 . PROBLEM 4 . * How to make ...
Seite 5
... draw A C , and A B , and you have the Angle required . Note , A right Angle is produced by laying off the Chord of 90 Degrees . Problem 5 . * Note . The quantity of Angles is determined by the parts of the Circumference of a Circle ...
... draw A C , and A B , and you have the Angle required . Note , A right Angle is produced by laying off the Chord of 90 Degrees . Problem 5 . * Note . The quantity of Angles is determined by the parts of the Circumference of a Circle ...
Seite 5
... Draw the Lines AC and A D , and you have the obtuse Angle required . PROBLEM 6 . To bifect a right lined Angle . 1. With any con- venient Radius in your Compaffes and one Foot in A , de- fcribe an Arch a b . A 2. With one Foot of your ...
... Draw the Lines AC and A D , and you have the obtuse Angle required . PROBLEM 6 . To bifect a right lined Angle . 1. With any con- venient Radius in your Compaffes and one Foot in A , de- fcribe an Arch a b . A 2. With one Foot of your ...
Seite 6
... Draw a ftraight Line through C and D , the Points of Interfection , which will cut the Line A B , into two equal Parts PROBLEM . 2 . To raise a perpendicular , at or near the Middle of a given right Line A B. gd , 1. From the given ...
... Draw a ftraight Line through C and D , the Points of Interfection , which will cut the Line A B , into two equal Parts PROBLEM . 2 . To raise a perpendicular , at or near the Middle of a given right Line A B. gd , 1. From the given ...
Andere Ausgaben - Alle anzeigen
An Introduction to Plain Trigonometry, with Its Application to Heights and ... Richard Cockrel Keine Leseprobe verfügbar - 2018 |
An Introduction to Plain Trigonometry, with Its Application to Heights and ... Richard Cockrel Keine Leseprobe verfügbar - 2016 |
Häufige Begriffe und Wortgruppen
A B C ABC are given alfo angled plain Triangle angled Triangle ABC Angles and Hypothenufe Arch D E Bafe and Hypothenufe becauſe bifect Cafe Centre Chord of 60 Compaffes confequently defcribe an Arch defcribe another Arch Diſtance draw A B fame fhall find the Angles find the Bafe find the Perpendicular firſt Foot fubtracted from 90 fuch fuppofe GEOMETRICALLY given right Line given the Hypothenufe gives the greater gives the leffer half the Difference half the Sum Houſe Hypoth Hypothenufe AC laſt lay off thereon lefs Line of Chords LOGARITHMETICALLY meaſured Number of Degs oblique angled plain oppofite Angle Parallelogram Perpend Perpendicu plain Triangle ABC Point of Interfection Proportion PROPOSITION Queſtion Radius 90 refpectively required the Angles right angled Triangle Scale of equal Secant Sect Side A B 70 Side and Angles Side BC 120 Sine Tang Tangent Trigono Turn to Plate
Beliebte Passagen
Seite 60 - As the base or sum of the segments Is to the sum of the other two sides, So is the difference of those sides To the difference of the segments of the base.
Seite 8 - If equal quantities be added to equal quantities, the sums will be equal.
Seite 75 - I measured a base of 500 yards in a straight line close by one side of it ; and at each end of this line I .found the angles subtended by the other end and a tree, close to the bank on the other side of the river, to be 53° and 79° 12'.
Seite 68 - What is the perpendicular height of a hill whose angle of elevation, taken at the bottom of it, was 46° ; and 100 yards farther off, on a level with the bottom of it, the angle was 31° ? Ans., 143.14 yards.
Seite 73 - I measured out for a base 400 yards in a right line by the side of the river, and found that the two angles, one at each end of this line, subtended by the other end and the house, were 68° 2
Seite 73 - Being on one side of a river and wanting to know the distance to a house, which stood on the other side, I measured 200 yards in a right line by the side of the river, and found that the two angles at each end of this line formed by the other end and the house were 73° 15' and 68° 2'; what was the distance between each station and the house ? Ans.
Seite 57 - As the sum of the two sides Is to their difference, So is the tan of half the sum of the unknown angles To the tan of half the difference of the unknown angles. And this tan half difference added to the half sum of two unknown angles gives the greater angle, and subtracted gives the less angle. Taking the example already given (fig. 29) : — AC + AD = 50 + 40...
Seite 17 - Triangles on the fame Bafe, and between the fame Parallels, are equal ; becaufe they are Half the circumfcribing Parallelograms.
Seite 70 - I meafured from its bottom a diftance of 40 feet, and then took the angle formed by the plane and a line drawn to the top 41"; and going 'on in the fame...
Seite 40 - DE, and ercdl the perpendicular DF ; which, it is evident, will be the tangent, and AF the fecant of the arc DE, or angle A, to the radius AD.