The sine of an angle is equal to the sine of its supplement, so that sin 147° 17' = sin 32° 43', whence log A C = 2.0681859 + 9.7327837-8.7645111 = 3.0364585 Therefore A C = 1087.6 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 A sin 3° 35' 30" whence log A D = log ll7 + L sin 87° 41' 30" - L sin 3° 35' 30" = 2.0681859 + 9.99964745 - 8.79688775 = 3.2709456 Therefore AD = 1866.15 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 now known, so that the triangles can be drawn upon the base A B by setting off the sides at the known angles, and the draughtsmanship can be checked by measuring the other known side of each triangle. The points C and D will then represent the positions of the two landmarks to which the observations were taken, and if the triangles 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 landmarks, the points A and B can be pricked through 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 magnitude of the angle A D C. Now, in any triangle the tangent 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'):: (1866.15 - 1087.6): (1866.15 + 1087.6), 778.55 tan 60° 20' therefore, tan 1/2 (ACD - ADC) = -------------------- 2953.75 or L tan 1/2 (ACD - ADC) = log 778.55 + L tan 60° 20' - log 2953.75 . = 2.8912865 + 10.2444l54 - 3.4703738 = 9.6653281 .•. 1/2 (ACD - ADC) = 24° 49' 53" .•. ACD - ADC = 49° 39' 46". Then algebraically (ACD + ADC) - (ACD - ADC) ADC = --------------------------- 2 120° 40' - 49° 39' 46" 71° 0' 14" .•. ADC = ------------------------- = ------------ = 35° 30' 7", 2 2 ACD = 180° - 35° 30' 7" - 59° 20' = 85° 9' 53". [Illustration: Fig. 35.--Arrangement of lines and Angles Showing Theodolite Readings and Dimensions.]