TO GEOMETRICAL PROBLEMS. 
305 
the whole being divided by r — z, their results r 4- 2$1* 
x r — z — f x r + ~z: which, ordered, gives 
3 rr — tt 
r 
X Z — x rr tt. 
4 
= «, and x rr — tt — /; then it 
4 
will be z 3 — az — f. Therefore find, from the tables, 
the arch whose co-sine is 
, (th e radius being 
unity); take 4 thereof, and find its co-sine; which, 
multiplied by 2s/\a, gives the true value .of'2 [see the 
last problem.) 
Thus, forHe^ample, let the radius OB (= r) zz 1, 
and the given *angle CD E rr 25°, whose tangent (t) is 
therefore — ,4663; whence %a rr ,23188, and \f zr 
,09782. 
Now, by logarithms, it will be log. \f — log. — 
i f 
\ log.-j-a rr — 1,9425328 rr log. -— L ^-=—rr log. co-sine 
20 t a\/ 
of 28° 50'; whereof the third part is 9 0 36f', whose 
log. co-sine (to the radius l) is—>1,9938609; which 
added to the \ log. of \a ( rr — 1,6826316)-gives 
— 1,6764925 rr log. of 0,47478, whose double ,94956 
is the true value of z, or FO : whence the correspond 
ing arch BE rr 18 J 16|', and consequently BC (rz2BE) 
rr 36° 33 ; .—By means of this problem that portion of 
a spherical surface representing the apparent figure 
of the sky is determined. 
4 PROBLEM XXXVI. 
The base AB,‘ and the difference of the angles at the 
base being given, while the angles themselves vary; to 
find the locus of the vertex E of the triangle. 
Let the base AB be bisected in O, and the angle BOD 
so constituted as to exceed its supplement AQD by the 
given difference of EAB and EBA ; and let ED, APQ, 
BSF be perpendicular, and EPF parallel to OD : 
then, since the angle BCE (BOD) as much exceeds 
x