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THE CONSTRUCTION OF 
BF and RF, whence the angle B is given; lastly, in thr 
triangle FQT will be given all the angles and the side 
FQ, whence QT and TF will he given, and conse 
quently PQ and FP. 
PROBLEM XXVI. 
To dir/de a given angle ABC into two parts CBF, 
ABF, so that their sines may obtain a given ratio. 
CONSTRUCTION. 
In BA, and CB produced, take BE and BD in the 
given ratio of the sine of CBF to the sine of ABF; 
draw DE, and parallel 
A. thereto draw BF, and 
the thing is done. For, 
by trigonometry, BE : 
BD : : the sine of D 
( zz CBF) : the sine of 
BED{-ABF). Hence 
the numerical solution is 
also evident : since it 
will be, as the sum of 
BE and BD is to their difference, so is the tangent of 
half the given angle ABC to the tangent of half the 
difference of the two required parts FBC and FBA. 
PROBLEM XXVII. 
To divide an angle given into two parts, so that their 
tangents may be to each other in a given ratio. 
CONSTRUCTION. 
Take any two right-lines AD, BD, which are in the 
ratio given,and upon the whole 
compounded line AB let a seg 
ment of a circle BCA be de 
scribed, capable of the angle 
given; make DC perpendi 
cular to AB, meeting the peri 
phery in C, and draw AC and 
BC, then will ACD and BCD 
be the two angles required.