A Treatise on Surveying,: Containing the Theory and Practice: : to which is Prefixed a Perspicuous System of Plane Trigonometry. : The Whole Clearly Demonstrated and Illustrated by a Large Number of Appropriate Examples. : Particularly Adapted to the Use of SchoolsKimber and Sharpless, no. 93, Market Street, and John Richardson, no. 244, Market Street., 1820 - 206 páginas |
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Página 26
... triangle is a figure bounded by three straight lines , as ABC , Fig . 6 . 15. An equilateral triangle has its three sides equal to each other , as A , Fig . 7 . 16. An isosceles triangle has only two of its sides equal , as B , Fig ...
... triangle is a figure bounded by three straight lines , as ABC , Fig . 6 . 15. An equilateral triangle has its three sides equal to each other , as A , Fig . 7 . 16. An isosceles triangle has only two of its sides equal , as B , Fig ...
Página 27
... triangle has one obtuse angle , as C , Fig . 10 . 20. An acute angled triangle has all its angles acute , as ABC , Fig . 6 . ! 21. Acute and obtuse angled triangles are called oblique angled triangles . 22. Any plane figure bounded by ...
... triangle has one obtuse angle , as C , Fig . 10 . 20. An acute angled triangle has all its angles acute , as ABC , Fig . 6 . ! 21. Acute and obtuse angled triangles are called oblique angled triangles . 22. Any plane figure bounded by ...
Página 28
... triangle . Thus AD is the base of the parallelogram ABEC , or triangle ABC , and CD is the altitude . Fig . 15 . 32. All plane figures contained under more than four sides , are called polygons ; of which those having five sides ...
... triangle . Thus AD is the base of the parallelogram ABEC , or triangle ABC , and CD is the altitude . Fig . 15 . 32. All plane figures contained under more than four sides , are called polygons ; of which those having five sides ...
Página 31
... ABC is the triangle required . PROBLEM IX . Upon a given line AB to describe a square , Fig . 26 . At the end B of the line AB , by problem 3 , erect the perpendicular BC , and make it equal to AB ; with A and Cas centres , and distance ...
... ABC is the triangle required . PROBLEM IX . Upon a given line AB to describe a square , Fig . 26 . At the end B of the line AB , by problem 3 , erect the perpendicular BC , and make it equal to AB ; with A and Cas centres , and distance ...
Página 32
... triangle ABC , Fig . 27 . By problem 1 , bisect any two of the sides , as AC , BC , by the perpendiculars DE and FG ; the point H where they intersect each other will be the centre of the circle ; with this centre , and the distance ...
... triangle ABC , Fig . 27 . By problem 1 , bisect any two of the sides , as AC , BC , by the perpendiculars DE and FG ; the point H where they intersect each other will be the centre of the circle ; with this centre , and the distance ...
Términos y frases comunes
ABCD ABFD acres adjacent adjacent angles angle ABC angle opposite angled triangle azimuth base bearing and distance breadth Calculation centre Co-secant Secant Co-sine Co-tang column decimal degrees DEMONSTRATION departure corresponding diff difference of latitude Dist divide division line draw equal EXAMPLES feet figures find the angle find the area four-pole chains fourth term given angle given area given bearing given ratio given side John Gummere Lat Dep latitude and departure length logarithm measured meridian distance natural number off-sets parallel parallelogram perches perpendicular perpendicular BC quired quotient radius rectangle remainder Required the area right angle right line right-angled triangle RULE semiperimeter side AB side AC sine square root station stationary line subtract tance Tang Tangent tract of land trapezium trapezoid triangle ABC trigonometry