Now join up points C and D on the plan, and from point D set off the line D A, making an angle of 35° 30' 7" with C D, and having a length of l866.15 ft, and from point C set off the angle A C D equal to 85° 9' 53". Then the line A C should measure l087.6 ft long, and meet the line A D at the point A, making an angle of 59° 20'. From point A draw a line A B, ll7 ft long, making an angle of 29° 23' with the line A C; join B C, then the angle ABC should measure 147° 17', and the angle B C A 3° 20'. If the lines and angles are accurately drawn, which can be proved by checking as indicated, the line A B will represent the base line in its correct position on the plan. The positions of the other stations can be calculated from the readings of the angles taken from such stations. Take stations E, F, G, and H as shown in Fig. 36*, the angles which are observed being marked with an arc. It will be observed that two of the angles of each triangle are recorded, so that the third is always known. The full lines represent those sides, the lengths of which are calculated, so that the dimensions of two sides and the three angles of each triangle are known. Starting with station E, Sin A E D: A D:: sin D A E: D E A D sin D A E D E = -------------- sin A E D or log D E = log A D + L sin D A E-L sin A E D. From station F, E and G are visible, but the landmark D cannot be seen; therefore, as the latter can be seen from G, it will be necessary to fix the position of G first. Then, sin E G D: D E :: sin E D G : E G, D E sin E D G or EG= --------------- sin E G D Now, sin E F G: E G :: sin F E G : F G E G sin F E G F G = ------------- sin E F G thus allowing the position of F to be fixed, and then sin F H G : F G :: sin F G H : F H F G sin F G H F H= ------------- sin F H G [Illustration: FIG 36.--DIAGRAM ILLUSTRATING TRIGONOMETRICAL SURVEY OF OBSERVATION STATIONS.] In triangles such as E F G and F G H all three angles can be directly read, so that any inaccuracy in the readings is at once apparent. The station H and further stations along the coast being: out of sight of landmark D, it will be as well to connect the survey up with another landmark K, which can be utilised in the forward work; the line K H being equal to F H sin K F H ------------- sin F K H The distance between C and D in Fig. 35 is calculated in a similar manner, because sin A C D : A D:: sin CAD : CD, AD sin CAD 1866.15 sin 59° 20' or CD = ---------- = ------------------- sin SCD sin 85° 9' 53" or log CD = log 1866.15 + L sin 59° 20' - L sin 85° 9' 53" = 3.2709456 + 9.9345738 - 9.9984516 = 3.2070678. ' . CD = 1610.90 ft