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other arch which is the measure of the fame angle, as the
radius of the first is to the radius of the second. Let AC, MN be measures of the angles ABC, according to
Def. 1. CD the fine, DA the versed fine, AE the tangent, and BE the secant of the arch: AC, according to Def. 4. 5. 6. 7. and NO the fine, OM the versed fine, MP the tangent, and BP the fecant of the arch MN, according to the fame definitions. Since CD, NO, AE, MP are parallel, CD is to NO as the radius CB to the radius NB, and AE, to MP as AB to BM, and BC or BA to BD as BN or BM to BO; and, by conversion, DA to MO as AB to MB. Hence the corollary is manifeft; therefore, if the radius be supposed to be divided into any given number of equal parts, the fine, versed fine, tangent, and secant of any given angle, will each contain a given number of these parts ; and, by trigonometrical tables, the length of the fine, versed sine, tangent, and secant of any angle may be found in parts of which the radius contains a given number: and, vice versa, a number expressing the length of the fine, versed sine, tangent, and secant being given, the angle of which it is the fine, versed fine, tangent, and secant
may be found.
plement of that angle. Thus, if BH be drawn perpendicular
and BK the secant of CBH, the complement of ABC, ac-
co-fine, and BK the co-fecant of the angle ABC. Cor. 1. The radius is a mean proportional between the tangent
and co-tangent. For, since HK, BA are parallel, the angles HKB, ABC will be
equal, and the angles KHB, BAE are right; therefore the triangles BAE, KHB are fimilar, and therefore AE is to AB,
as BH or BA to HK. Cor. 2. The radius is a mean proportional between the co-fine
and fecant of any angle ABC. Since CD, AE are parallel, BD is to BC or BA, as BA to BE.
PROP. I. Na right angled plain triangle, if the hypothenuse
be made radius, the sides become the fines of the angles opposite to them; and if either side be made radius, the remaining fide is the tangent of the angle opposite to it, and the hypothenuse the fecant of the fame angle.
Let ABC be a right angled triangle; if the hypothenuse BC be made radius, either of the sides AC will be the fine of the angle ABC opposite to it; and if either side BA be made radius, the other side AC will be the tangent of the angle ABC opposite to it, and the hypothenuse BC the secant of the same angle.
About B as a center, with BC, BA for distances, let two circles CD, E A be described, meeting BA, BC in D, E: since CAB is a right angle, BC being radius, AC is the fine of the angle ABC by def. 4. and BA being radius, AC is the tangent, and BC the fecant of the angle ABC, by def. 6. 7.
Cor. 1. Of the hypothenuse a side and an angle of a right angled triangle, any two being given, the third is also given.
Cor. 2. Of the two sides and an angle of a right angled triangle, any two being given, the third is also given.
as the fines of the angles opposite to them. In right angled triangles this prop. is manifest from prop. 1. for if the hypothenuse be made radius, the sides are the fines of the angles opposite to them, and the radius is the fine of a right angle (cor, to def. 4.) which is opposite to the hypothenuse.
In any oblique angled triangle ABC, any two fides AB, AC will be to one another as the lines of the angles ACB, ABC which are opposite to them.
From C, B draw CE, BD perpendicular upon the opposite Gdes AB, AC produced, if need be. Since CEB, CDB are right angles, BC being radius, CE is the fine of the angle CBA, and BD the fine of the angle ACB; but the two triangles CAE, DAB have each a right angle at D and E; and likewise the common angle CAB; therefore they are similar, and consequently,
FIG. 6. 7.
CA is to AB, as CE to DB; that is, the sides are as the fines of the angles opposite to them.
Cor. Hence of two fides, and two angles oppofite to them, in a plain triangle, any three being given, the fourth is also given.
PROP. III. Fig. 8. N a plain triangle, the sum of any two fides is to
their difference, as the tangent of half che sum of the angles at the base, to the tangent of half their difference.
Let ABC be a plain triangle, the sum of any two sides AB, AC will be to their difference as the tangent of half the fum
of the angles at the base ABC, ACB to the tangent of half 'their difference.
About A as a center, with AB the greater side for a distance, let a circle be described, meeting AC produced in E, F, and BC in D; join DA, EB, FB; and draw FG parallel to BC, meeting EB in G.
The angle EAB (32. 1.) is equal to the sum of the angles at the base, and the angle EFB at the circumference is equal to the half of EAB at the center (20. 3.); therefore EFB is half the sum of the angles at the base; but the angle ACB (32. 1.) is equal to the angles CAD and ADC, or ABC together; therefore FAD is the difference of the angles at the base, and FBD at the circumference, or BFG, on account of the parallels FG, BD, is the half of that difference; but since the angle EBF in a semicircle is a right angle (1. of this) FB being radius, BE, BG, are the tangents of the angles EF 5, BFG; but it is manifest that EC is the sum of the fides BA, AC, and CF their difference; and since BC, FG are parallel (2. 6.) EC is to CF, as EB to BG; that is, the sum of the sides is to their difference, as the tangent of half the sum of the angles at the base to the tangent of half their difference.
N any plane triangle BAC, whose two sides are BA,
AC and base BC, the less of the two fides, which let be BA, is to the greater AC as the radius is to the tangent of an angle; and the radius is to the
tangent of the excess of this angle above half a right angle as the tangent of half the sum of the angles B and C at the base, is to the tangent of half their dif. ference.
At the point A, draw the straight line EAD perpendicular to BA; make AE, AF, each equal to AB, and AD to AC; join BE, BF, BD, and from D, draw DG perpendicular upon BF. And because BA is at right angles to EF, and EA, AB, AF are equal, each of the angles EBA, ABF is half a right angle, and the whole EBF is a right angle; also (4. 1. El.) EB is equal to BF. And since EBF, FGD are right angles, EB is parallel to GD, and the triangles EBF, FGD are similar ; therefore EB is to BF as DG to GF, and EB being equal to BF, FG must be equal to GD. And because BAD is a right angle, BA the less fide is to AD or AC the greater, as the radius is to the tangent of the angle ABD; and because BGD is a right angle, BG is to GD or GF as the radius is to the tangent of GBD, which is the excess of the angle ABD above ABF half a right angle. But because EB is parallel to GD, BG is to GF as ED is to DF, that is, fince ED is the sum of the sides BA, AC and FD their difference, (3. of this,) as the tangent of half the sum of the angles B, C, at the base to the tangent of half their difference. Therefore, in any plain triangle, &c. Q. E. D.
PROP. V. Fig. 9. and 10. N any triangle, twice the rectangle contained by
any two sides is to the difference of the sum of the squares of these two sides, and the square of the base, as the radius is to the co-line of the angle in: cluded by the two sides.
Let ABC be a plain triangle, twice the rectangle ABC contained by any two sides BA, BC is to the difference of the sum of the squares of BA, BC, and the square of the base AC, as the radius to the co-fine of the angle ABC.
From A, draw AD perpendicular upon the opposite fide BC; then (by 12. and 13. 2. El.) the difference of the sum of the squares of AB, BC, and the square of the base AC, is equal to twice the rectangle CBD; but twice the rectangle CBA is to
twice the rectangle CBD, that is, to the difference of the sum of the squares of AB, BC, and the square of AC, (1. 6.) as A3 to BD; that is, by prop. 1. as radius to the fine of BAD, which is the complement of the angle ABC, that is, as radius to the Co-fine of ABC.
PROP. VI. Fig. 11. N any triangle ABC, whose two sides are AB, AC,
and base BC; the rectangle contained by half the perimeter, and the excess of it above the base BC, is to the rectangle contained by the straight lines, by which the half of the perimeter exceeds the other two fides AB, AC, as the square of the radius is to the square of the tangent of half the angle BAC opposite to the base.
Let the angles BAC, ABC be bifected by the straight lines AG, BG; and, producing the fide AB, let the exterior angle CBH be bisected by the straight line BK, meeting AG in K; and from the points G, K, let there be drawn perpendicular upon the sides the straight lines GD, GE, GF, KH, KL, KM. Since therefore (4. 4.) G is the center of the circle inscribed in the triangle ABC, GD, GF, GE will be equal, and AD will be equal to AE, BD to BF, and CE to CF. In like manner KH, KL, KM will be equal, and BH will be equal to BM, and AH to AL, because the angles HBM, HAL are bifected by the straight lines BK, KA: and because in the triangles KCL, KCM, the fides LK, KM are equal, KC is common and KLC, KMC are right angles, CL will be equal to CM: since therefore BM is equal to BH, and CM to CL; BC will be equal to BH and CL together; and, a:lding AB and AC together, AB, AC, and BC will together be equal to AH and AL together : but AH, AL are equal : wherefore each of them is equal to half the perimeter of the triangle ABC: but since AD, AE are equal, and BD, BF, and also CE, CF, AB together with FC, will be equal to half the perimeter of the triangle to which AH or AL was shewn to be equal; taking away therefore the common AB, the remainder FC will be equal to the remainder BH: in the same manner is it demonstrated, that BF is equal to CL, and since the points B, D, G, F, are in a circle, the angle DGF will be