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THEOREM I. .
In any right-angled plane triangle ABC, it will be ate the bypothenuse is to the perpendicular, so is the radius of the table) to the line of the angle at the base.,
G For, let AE or AF E
be the radius to which the table of lines, &c. is adapted, and ED the line of the angle A or arch EF (l'id.
Def. 3. and 6.); then, B D F because of the fimi
lar triangles ACB and AED, it will be AC: BC::AE:ED (by 14:4.) Q: E. D.
Thus, if AC = 975, and BC = 245; then it will be, ,75 : ,45 :: 1 (radius): the line of A =,6; which, in the table, answers to 36° 52', the measure, or value of A.
THEOREM II. In any right-angled plane triangle ABC, it will be, as the base AB is to the perpendicular BC, so is the radius of the table) to the tangent of the angle af the base.
For, let AE or AF be the radius of the table, or canon (see the preceding figure), and FG the tangent of the angle A, or arch: EF (Vid. Def. 3. and 9.); then, by reason of the similarity of the triangles ABC, AFG, it will be, AB : BC : : AF: FG. 2: E. D.
Note, In the quotations where you meet with two numbers (as 14. 4 ) without any mention of Prop to the Elements of Geometry published by tbe fame author; to ebicbihis little tra& is designed as an Appendix.
Thus let AB =,8, and BC =,5; then we shall have , 8 : ,5 :: 1 (radius) : tangent A =,625; whence A itself is found, by the canon, to be
In every plane triangle ABC, it will be, as any one fide is to the fine of its opposite angle, fo is any other fide to the fine of its opposite angle.
For take CF = B AB, and upon AC let fall the perpen
F diculars BD and FE; which will be the lines of the angles A and C to the equal radii AB and CF.A D Now the triangles CBD,'CFE being similar, we have CB : BD (fin. A) :: CF (AB): FE (fin. C). 2. E. D.
THEOREM IV. As the base of any plane triangle ABC, is to the fum of the two sides, so is the difference of the fides to twice the distance DE of the perpendicular from the middle of the base.
For (by Cor. to 9. 2.) AB + BC XAB-BC = ACX
2DE; whence AC : AB + BC :: AB BC: 2DE (by 10. 4.) 2. E. D.
THÉOR E MV.: In any plane triangle, it will be, as the sum of any two sides is to their difference, so is the tangent of half the sum of the two opposite angles, to the tangent of balf their difference.
For, let ABC be the triangle, and
AB and AC the А.
two proposed fides; F
and from the cenD
ter A, with the vadius AB, let a circle be described, in.
tersecting CA pro6
dụced, in D and F;
so that CF may express the sum, and CD the difference, of the sides AC and AB : join F, B and B, D, and draw DE parallel to FB, meeting BC in E.
Then, because 2ĄDB = ADB + ABD (by 12. 1.) = C + ABC (by 9. 1.) it is plain that ADB is equal to half the sum of the angles opposite to the fides proposed, Moreover, since ABC = ABD (ADB) + DBC, and C = ADB DBC (by 9. 1.) it is plain that ABC - Ciş = 2DBC ; or that DBC is equal to half the difference of the fame angles.
Now, because of the parallel-lines BF and ED, it will be CF:CD::BF: DE, but BF and DE, because DBF and BDE are right-angles (by: 13.3: and 7. 1.) will be tangents of the foresaid angles FDB (ADB) and DBE (DBC) to the radius BD. 2. E. D.
Hence, in two triangles ABC and ABC, having two sides equal, each to each, it will be (by equa
ABC +ACb ABC-AC) lity), as tang
: tang ABC + ACB ABC-ACB tang. : tang
But, if CAb be supposed a right-angle, then will AbC+ ACb allora right angle (by Cor. 3. to 10. 1.) and
ALC + ACB the tangent of
– radius. Therefore in this case our proportion will become,
АС АСь Radius : tang.
(=ABC - 45°):: ABC+ACB ABC-ACB tang
Which gives the following Theorem, for finding the angles opposite to any two proposed sides, the in- . cluded angle, and the sides themselves, being known.
As the lefser of the proposed fides (Ab or AB) is to the greater (AC), jo iş radius to the tangent of an angle (AbC, see Theor. 2.) And as radius to the tangent of the excess of this angle above 45°, so is the tangent of half the sum of the required angles to the tangent of half their difference*.
• This Theorem, though it requires trvo proportions, is commonly used by Aftronomers in determining the elongation and parallaxes of the planets (being best adapted to logarithms); for wohich reason is is here given.
The solution of the cases of right-angled plane triangles.
Solution. 1 he hyp. Une leg As radius is to the line of I AC and
BC A, so is the hyp. AC to the angles
che leg BC (by Theor. I.) The hyp. The an- As AC : BC :: radius : 2 AC and
gles lin. A (Théor. 1.) whose one legBC
complement is the angle C. The hyp. The other Ler the angles be found, 3 AC and leg AB by Case 2. and then the reone legBC quired leg AB. by Cafe 1.
The an- The, hyp. As fine A : radius : : the 4 gles and AC leg BC:
to the hyp. one legBC
AC (Theor. 1.) The an- The other As fine A: BC :: fine C gles and leg AB : AB (by Theor. III.) Or, one leg
jas radius : tang. C :: BC BC
: AB (by Theor. II.) The two
The an- As AB : BC :: radius 6
tang, A (by Theorem 11.) and BC
whole complement is the
angle C.. The two The hyp. Let the angles be found, 7 legs AB
AC by Cafe 6. and then the and BC
hyp. AC, by Case 4.