GANNETT. 1 
Example: 
SOLUTION OF TRIANGLES. 
(liven log a=4. 3666779 Given C (spherical angle) 21° 14' 54".10 
Given log b=4. 2050498 Given £ sph. exc. - .10 
13 
(1) tan x= 0.1616281 
z=55° 25' 25. 11" 
-45 
(5) Logtan (.x-45°)=10° 25' 25". 41=9. 2647291 
(6) Log tan 79 22 33 .00=0.7268100 
C (plane angle) = 21 14 54 .00 (2) 
180 
180°-C=A+B=158 45' 06 .00 (3) 
a (A+B)= 79° 22' 33". 00 (4) 
(7) 
9. 9915391= tan i (A-B) 
44 26 30 .90 
^Hm--A=123° 49' 03". 90 (8) 
difference=B= 34 56 02 .10 (9) 
(10) 
Check. 
A=123° 49' 03". 90 
B= 34 50 02 .10 
C= 21 14 54 .00 
log a =4. 3666779 
a. c. log sin A=0. 080-1 971 
log sin B=9. 7578749 
log sin C=9. 5592012 
Sum = 180 00 00 .00 
log c =4. 0063762 
log b =4. 2050499 
TimEE-POHSTT PROBLEM. 
If three points, forming a triangle of which the sides and angles are 
known or can be computed, be visible from a fourth point, P, it is 
required to determine the position of P. 
Set up the theodolite at P and measure the two angles subtended by 
any two of the given sides. 
This problem is of use in cases where, the regular triangulation hav- 
ing been completed, additional points are required for the topographic 
survey, or are needed for special service. The angles should be care- 
fully measured, and in the computations the logarithms should be car- 
ried to seven places of decimals. 
Three cases of its application are given, as in others, such as when 
P falls upon one or the other of the sides of the known triangle, or on 
the prolongation of either, the case resolves itself into the solution of 
a simple triangle with one side and the angles given; or the problem is 
indeterminate, as when P is situated on the circumference of the circle 
passing through the three known points — a contingency which rarely 
occurs. 
