fore the angle C AE^ L CFE=^CFH= (by conftrudion) 
to the obferved angle y c^g. la like manner the ^ CE A~ 
CFJ=CFG^' obierv'd angle y id. 
If the ftations A, E fall in a right line with the 
point C , the lines G 4 H being parallel, cannot 
meet: but in this cafe the problem is indeterminate 
and capable ot infinite Solutions. For as before upon 
CG delcnbe a Segment of a circle capable of the ob- 
ferved I yict,, and upon C t-', defcribe a Segment capable 
of the obferved >ctg : then thro draw a line any way 
cutting the circles in A,E, thefe points wilanlwer the 
queftion. 
The Third Prollem. 
Four points B, C, D, F, (Fig. 6.) or the 4 fides of a qua- 
drilateral, with the angles comprehended are given ; 
alfo there are 2 ftations A and E fuch, that at A, only 
B,C,E are vifible, and at E only AyD.P, that is, the 
angles B^f, BAE, AED, DEF arc given: to find 
the places of the two points A,Ei and confequently, 
the lengthsof the lines A3, AC, AE, ED, EF. 
Conftru^lion. 
Upon BC (by 33. 3.Eucl. ) de/cribe a fegnientof a 
circle, that may contain an angle equal to the obferved 
angle 5 then from f draw the chord CM, or a line 
cutting the circle in M, fo that the angle BCM may 
be equal to the lupplement of the obferved angle BAE, 
i. e. Its refidue to 180 degrees. In like manner on D f 
defcribe a legment of a circle, capable of an angle equal 
to the obferv'd D EF, and from D draw the chord Z> iV, 
fo that the angle FD iVmay be equal to the fupplement 
of the obferv'd angle joiiiMW, cuttingthe 2 cir- 
cles in Ay E '. I fay A, E^ are the two points requir'd. 
Dem : 
Join ^i?, AC, ED, EF, then is the I MAB = ^ 
BCM{bj 21. 3. Eucl:)=fupp;ementof the- obferv'd 
by conftrudion, therefore the conftrufited 
iBAE 
