CHAP. XV.] TRUE BEARING 357 



or more from the sun, these conditions will be fulfilled, the sun 

 being, of course, comparatively low, and near the prime vertical. 



There are two separate cases : — Two 



First, when the object whose bearing is desired is on the ^*^®^' 

 horizon ; and secondly, when it has a sensible altitude, as a 

 mountain top. 



In the first we have to solve a quadrantal triangle as shown 

 in Fig. 70. 



In this Fig. Z is zenith, S is sun, and the object on the object on 

 horizon. Horizon. 



We have Z O = 90°. Z S the apparent zenith distance, and 

 S the observed angular distance, to find O Z S, the horizontal 

 angle required, or 



Cos horiz. angle = Cos ang. dist. x Sec app. alt. 

 with the proper signs applied to the angles. 



Example. 



(Object on horizon, two observers with sextants and sea 



horizon.) 



On June 1, 1881, at Ship IV. 7 hours 24 minutes a.m. mean 

 time of place, observed altitude of Q 30° 13' 50", mean angular 

 distance of C| to Pine A on horizon 84° 26' 20", object right of 

 Lat. 40° 26' 15" N., Long. 28° 00' E. Index Errors -35" 

 and 0". H.E. 20 feet. 



b. m. o / // 



M. Time pi. .. 7-24 ©'s dec. 1st .. 22 6 59-6 N. 



Long, in time . . 1 • 52 2 03 



Gr. Date 31st .. 17-32 Corrected dec. . . 22 04 51-6 



1st .. - 6-28 Pol. dis. .. 67 55 03 



a I II I, 



Obd. alt. .. 30 13 £0 Yar. .. 20-0 



Index error . . — 35 6 • 4 



800 

 H.E. .. 4 15 1200 



S. D. 



App. alt. . . 30 24 48 Obs. Ang. dist. 



Eef. .. -1 31 S. D. 



T. alt. ,. 30 23 17 True An?, dist. 



