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 |
Dentro del libro
Página 71
... meeting CA produced , in D : then AD , CD and BD will be the distances required . * Calculation . 1. In the triangle ABC we have all the sides given , to find the angle C 104 ° 8 ' . = 2. Subtract the sum of the angles D and C from 180 ...
... meeting CA produced , in D : then AD , CD and BD will be the distances required . * Calculation . 1. In the triangle ABC we have all the sides given , to find the angle C 104 ° 8 ' . = 2. Subtract the sum of the angles D and C from 180 ...
Página 72
... meeting in E , and ( by prob . 10. ) describe a circle that shall pass through the points A , E and B : join CE and produce it to meet the circle in D , and join AD , BD , then will AD , CD and BD be the distances required ...
... meeting in E , and ( by prob . 10. ) describe a circle that shall pass through the points A , E and B : join CE and produce it to meet the circle in D , and join AD , BD , then will AD , CD and BD be the distances required ...
Página 78
... meeting BC in E ; then AECE = the part broken off . * Calculation . 1. In the right - angled triangle ABC , we have AB = 34 and BC = 100 , to find the angle C = 18 ° 47 ′ . 2. In the right angled triangle ABE , we have AEB = ACE + CAE ...
... meeting BC in E ; then AECE = the part broken off . * Calculation . 1. In the right - angled triangle ABC , we have AB = 34 and BC = 100 , to find the angle C = 18 ° 47 ′ . 2. In the right angled triangle ABE , we have AEB = ACE + CAE ...
Página 93
... meeting AG * pro- duced in K , join KC , and let fall the perpendiculars KH , KM , and KL . Then ( 26.1 ) AD is equal to AE and DG to GE ; also BD is equal to BF and DG to GF ; hence GF and GE are equal , and conse- quently ( 47.1 ) CF ...
... meeting AG * pro- duced in K , join KC , and let fall the perpendiculars KH , KM , and KL . Then ( 26.1 ) AD is equal to AE and DG to GE ; also BD is equal to BF and DG to GF ; hence GF and GE are equal , and conse- quently ( 47.1 ) CF ...
Página 97
... meeting BC in C ; lastly draw CD parallel to AF , meeting AD in D , then will ABCD be the trapezium . * Calculation . The angle E = 180 ° the sum of the angles BCD , ADC23 ° 20 ' . * DEMONSTRATION . By construction FC is parallel to AD ...
... meeting BC in C ; lastly draw CD parallel to AF , meeting AD in D , then will ABCD be the trapezium . * Calculation . The angle E = 180 ° the sum of the angles BCD , ADC23 ° 20 ' . * DEMONSTRATION . By construction FC is parallel to AD ...
Términos y frases comunes
ABCD ABFD acres adjacent adjacent angles angle ABC angle opposite angled triangle base bearing and distance breadth Calculation centre Co-secant Secant Co-sine Co-tang column cube root decimal 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 number given ratio given side John Gummere Lat Dep latitude and departure length logarithm M.
M. Sine measured meridian distance natural number off-sets parallel parallelogram perches perpendicular quired quotient radius rectangle remainder Required the area right angle right line right-angled triangle RULE semiperimeter side AB side AC square root station stationary line subtract survey tance Tang tangent tract of land trapezium trapezoid triangle ABC trigonometry