Mr B. Martin’s Experiments on Island Crystal. 151 
From the angular point B let fall a perpendicular BI to the 
ground-plane, and through I draw KM indefinitely ; then will 
the angle BKI be 70° 57, the obliquity of the edge or line BK 
of the solid ; and its complement the angle IBK will be 19° 3'. 
Let h K be perpendicular to the side BC, as also the line hi; 
then the angle ihK was found by exact measurement to be 105 
degrees, which is the measure of the inclination of the planes 
ACBD and CBKF to each other. 
Let a plane CGHF perpendicular to the horizon, be supposed 
to cut the solid through the side or edge CF, then will it 
evidently bisect the obtuse angle ACB, and so make the angles 
ACG, BCG, of 50° 56' each. Also it is evident, the lines KM 
and FH will be parallel. Draw the diagonal CH. Then our illus- 
trious author proceeded to computation in the following manner. 
Let C be the vertex of a triangular pyramid (in the centre 
of a sphere) contained under three equal and equi-angular planes 
ACB, ACF, BCF, (Fig. 2.), then will the base of the pyramid 
be the spherical triangle ABF, equilateral and equiangular. 
And then it is plain, that each angle as A measures or is equal 
to the inclination of the two containing planes ACB and ACF ; 
and each side of the triangle as AB is the measure of the oppo- 
site angle ACB. And further, if any angle as ACB be bisected by 
a line C G, it will of course also bisect the side AB, and drawing 
the arch FG, it will bisect the angle at F. In like manner, if 
the side B F be bisected in O, the arch A O bisects the angle at 
A, and cuts the other arch FG in H, the pole of the sphere ; 
and CH is the semiaxis thereof. Lastly, suppose the three plain 
angles which compose the solid angle C in each figure the same. 
Then because AFB is equal to IhK (in Fig. I.) or 105 
degrees, the half thereof AFG will be 52° SO', and the angles 
at G being right ones, all the angles of the right angled triangle 
AFG, or BFG, are known ; and therefore the side AF, or BF, 
will be found to be 101° 52 ', which is, therefore, the measure or 
quantity of each obtuse angle forming the solid angles C and E 
in the Island Crystal. 
In the same triangle AFG, the side or hypothenuse F G is 
found to be 109° 3', and as this is the measure of the angle FCG 
(Fig. 1.), so its complement to 180°, viz. 70° 57' will be the quan- 
tity of the angle CFH or BKM in the crystal ; and, therefore, 
the angleJLBK will be 19° 3', as we have said. 
