Part 10 (1/2)

Chapter XIV

TRIGONOMETRICAL SURVEYING

In the surveying work necessary to fix the positions of the various stations, and of the float, a few eleonoeously explained by taking practical exa selected the th of any line A B on a convenient piece of level ground, the next step will be to fix its position upon the plan Two pro-staffs, etc, the positions of which are shown upon the ordnance les read from each of the stations A and B assuularfrom zero on that line are, from A to point C, 29 23' and to point D 88 43', and from B to point C 212 43', and to point D 272 18' 30” The actual readings can be noted, and then the arrange 35, froths AC and AD As the three angles of a triangle equal 180, the angle B C A = 180- 147 17'-29 23'= 3 20', the angle B D A = 180-87 41' 30”- 88 43'= 3 35' 30” In any triangle the sides are proportionate to the sines of the opposite angles, and vice versa; therefore,

A B : A C :: sin B C A : sin A B C, or sin B C A : A B :: sin ABC : A C, nr A C = (A B sin A B C) / (sin B C A) = (117 x sin 147 17') / (sin 3 20')

or log A C = log 117 + L sin 147 17' - L sin 3 20'

The sine of an angle is equal to the sine of its supple A C = 20681859 + 97327837-87645111 = 30364585

Therefore A C = 10876 feet

Similarly sin B D A: A B :: sin A B D: A D

A B sin A B D 117 x sin 87 41' 30”

therefore A D = --------------- = ----------------------- sin B D Asin 3 35' 30”

whence log A D = log ll7 + L sin 87 41' 30” - L sin 3 35' 30”

= 20681859 + 999964745 - 879688775 = 32709456

Therefore AD = 186615 feet

The length of two of the sides and all three angles of each of the two triangles A C B and A D B are non, so that the triangles can be drawn upon the base A B by setting off the sides at the known angles, and the draughts the other known side of each triangle The points C and D will then represent the positions of the two landles are drawn upon a piece of tracing paper, and then superimposed upon the ordnance map so that the points C and D correspond with the landh on to the map, and the base line A B drawn in its correct position

If it is desired to draw the base line on the map direct from the two known points, it will be necessary to ascertain the ent of half the difference of two angles is to the tangent of half their sum as the difference of the two opposite sides is to their sum; that is:--

Tan 1/2 (ACD - ADC): tan 1/2 (ACD + ADC):: AD - AC : AD + AC,

but ACD + ADC = l80 - CAD = 120 40', therefore, tan 1/2 (ACD - ADC): tan 1/2 (120 40'):: (186615 - 10876): (186615 + 10876),

77855 tan 60 20'

therefore, tan 1/2 (ACD - ADC) = -------------------- 295375

or L tan 1/2 (ACD - ADC) = log 77855 + L tan 60 20'

- log 295375

= 28912865 + 102444l54 - 34703738 = 96653281 ? 1/2 (ACD - ADC) = 24 49' 53”

? ACD - ADC = 49 39' 46” Then algebraically