Trigonometrical Survey. 627 



In Fig. 4. Plate XXX. let Z be the zenith, B the station on 

 Black Down, and ZBA its meridian ; also, let D be Dun nose, 

 and ZD its meridian ; likewise, suppose BC to be an arc of a 

 great circle, perpendicular to the meridian at B, and DA ano- 

 ther arc of a great circle, perpendicular to the meridian at D, 

 BF and ED being the parallels of latitude at Black Down and 

 Dunnose. 



In the spherical triangle BZD, the angles at B and D are 



given, the first being 94 2' 22", J5, and the second 84 54' 53"; 



therefore, in the triangle ABD the angle at B is 85° 57' 36",75, 



and, in the triangle BDC, the angle at D = 84,° 54' 53" : hence, 



the angles of these triangles, when reduced to those formed by 



the chords, are as follows : 



rDDC = 84°54/52,5" 

 In the triangle BDC \ CDB = 91 2 44,75 



lCBD= 4 2 22,75 



f ABD = 85 57 36,75 

 And in the triangle ABDJ BAD = 88 5 y 16,25 



lBDA= 5 5 7 

 Now the distance between Black Down and Dunnose, BD, 

 has been already found to be 314307,5 feet; therefore, using 

 the above angles with that distance, (after the proper corrections 

 are applied for reducing the horizontal angles to those formed 

 by the chords,) we get, 



In the triangle BCD ( J*£ = 3*3 l2 ? 1 f t 

 & [CD = 21146,9 J 



And in the triangle ABD f^? = 31 §£ 8l ' 2 lfeet. 

 & \AB = 27864,5 J 



Again, in the two small triangles formed by the parallels 



BF and ED, the perpendiculars BC and DA, and the small 



arcs CF and AE, we have the angles at C and A given, the 



