270 



DEEP BOREHOLE SURVEYS AND PROBLEMS 



To get the full dip 5, applying the rule of Eq. (52) above 

 we get 



tan d sin 90' 



or tan 5 = 



tan (S 



(56) 



tan /? sm ^o sm 02 



so that Eqs. (55) and (56) provide the full solution. 



Second Solution. — This alternate method is adopted 



when we do not desire to obtain the angles di and 62 by 



construction. Let the 

 known lines ABi and BE 

 equal h and U, also let the 

 known depth difference 

 between B and A he hi 

 and between C and B be 

 /i2. Produce ABi (Fig. 

 179) to take a perpendic- 

 ular DG let fall on it from 



D, and draw BFi and BiFi meeting at Fi each normal to 



AD the strike line so that the angle BFiBi = 5 the full dip. 



DBi = hi cot /? (57) 



DG 



Fig. 179. 



tan 4>2 = 

 DG - 



GBi + BiA 



DBi sin 01 



(58) 



and 



GBi = DBi cos 4>i 

 By substituting in Eq. (58) we get 



_ DBi sin <pi _ hi cot j8 sin 0i 

 tan <p2 — 



tan 02 = 



(59) 



/ll?2 COS 01 + /I2/1 



Third Solution. — Using the above figure and notation 

 and noting that the bearing of AC is /3i, of BC, ^2 and for 



