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the tangent of half the sum of the angles at the base, to the tangent of half their difference.

PROP. IV. FIG. 18. In any plane triangle BAC, whose two sides are BA, AC, and base BC, the less of the two sides, 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 difference.

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 ÉBA, 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 side 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 GĎ 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 EĎ is parallel to GD, BG is to GF, as ED is to DF; that is, since 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 plane triangle, &c. Q. E. D.

PROP. V. Fig. 9. and 10. In 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 cosine of the angle included by the two sides.

Let ABC be a plane triangle, twice the rectangle ABD 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 cosine of the angle ABC.

From A, draw AD perpendicular upon the opposite side BC, then (by 12. and 13. 2. EI.) 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 AB (1.6.), as AB to BD; that is, by Prop. 1. as radius to the sine of BAD, which is the complement of the angle ABC; that is, as radius to the cosine of ABC..

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PROP. VI. Fig. 11. IN 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 sides 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 bisected by the straight lines AG, BG; and producing the side 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 centre 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 bisected by the straight lines BK, KA: And because in the triangles KCL, KCM, the sides LK, KM are equal, KC is common, and KLC, KMC are right angles, CL will be equal to ĆM: Since therefore BM is equal to BH, and CM to CL; BC will be equal to BH and CL together; and, adding AB and AC together, AB, AC, and BC will together be equal to AH and Aų 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

shown to be equal; taking away therefore the common AB, the remainder FC will be equal to the remainder BH: In the same manner it is demonstrated, that BF is equal to CL: And since the points B, D, G, F, are in a circle, the angle DGF will be equal to the exterior and opposite angle FBH (22. 3.); wherefore their halves BGD, HBK, will be equal to one another: The right angled triangles BGD, HBK, will therefore be equiangular, and GD will be to BD, as BH to HK; and the rectangle contained by GD, HK, will be equal to the rectangle DBH or BFC: But since AH is to HK, as AD to DG, the rectangle HAD (22.6.) will be to the rectangle contained by HK, DG, or the rectangle BFC, (as the square of AD is to the square of DG, that is) as the square of the radius is to the square of the tangent of the angle DAG, that is, the half of BAC: But HA is half the perimeter of the triangle ABC, and AD is the excess of the same above HD, that is, above the base BC; but BF or CL is the excess of HA or AL above the side AC, and FC, or HB, is the excess of the same HA above the side AB; therefore the rectangle contained by half the perimeter, and the excess of the same above the base, viz. the rectangle HAD, is to the rectangle contained by the straight lines by which the half of the perimeter es. ceeds the other two sides, that is, the rectangle BFC, as the square of the radius is to the square of the tangent of half the angle BAC opposite to the base. Q.E.D.

PROP. VII. Fig. 12, 13. In a plane triangle the base is to the sum of the sides as the difference of the sides is to the sum or difference of the segments of the base made by the perpendicular upon it from the vertex, according as the square of the greater side is greater or less than the sum of the squares of the lesser side and the base.

Let ABC be a plane triangle; if from A the vertex be drawn a straight line AD, perpendicular upon the base BC, the base BC will be to the sum of the sides BA, AC, as the difference of the same sides is to the sum or difference of the segments CD, BD, according as the square of AC the greater side is greater or less than the sum of the squares of the lesser side AB, and the base BC.

About A, as a centre, with AC the greater side for a distance, let a circle be described meeting AB produced in E, F, and CB in G: It is manifest, that FB is the som, and BE

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the difference of the sides; and since AD is perpendicular to GC, GD, CD will be equal; consequently GB will be equal to the sum or difference of the segments CD, BD, according as the perpendicular AD meets the base produced, or the base; that is (by 12. 13. 2.), according as the

square of AC is greater or less than the sum of the . squares of AB, BC: But (by 35. 3.) the rectangle CBG is

equal to the rectangle EBF; that is (16.6.) BC is to BF, as BE is to BG; that is, the base is to the sum of the sides as the difference of the sides is to the sum or difference of the segments of the base made by the perpendicular from the vertex, according as the square of the greater side is greater or less than the sum of the squares of the lesser side and the base. Q. E. D.

PROP. VIII. PROB. Fig. 14. The sum and difference of two magnitudes being given, to find them.

Half the given sum added to half the given difference, will be the greater, and half the difference subtracted from half the

sum,

will be the less. For let AB be the given sum, AC the greater, and BC the less. Let AD be half the given sum; and to AD, DB, which are equal, let DC be added; then AC will be equal to BD and DC together; that is, to BC, and twice DC; consequently, twice DC is the difference, and DC half that difference; but AC the greater is equal to AD, DC; that is, to half the sum added to half the difference, and BC the less is equal to the excess of BD, half the sum,

above DC half the difference. Q.E.D.

SCHOLIUM. Of the six parts of a plane triangle (the three sides and three angles) any three being given, to find the other three is the business of plane trigonometry; and the several cases of that problem may be resolved by means of the preceding propositions, as in the two following, with the tables annexed.

In these, the solution is expressed by a fourth proportional to • three given lines; but if the given parts be expressed by

numbers from trigonometrical tables, it may be obtained arithmetically by the common Rule of Three.

Note. In the tables the following abbreviations are used : R. is put for the Radius ; T. for Tangent; and S. for Sine. Degrees, minates, seconds, &c. 1. are written in this manner: $0° 25' 13", &c. which signifies 30 degrees, 25 minutes, 13 seconda de

Solution of the Cases of Right Angled Triangles.

GENERAL PROPOSITION. In a right angled triangle, of the three sides and three angles, any two being given besides the right angle, the other three may be found, except when the two acute angles are given; in which case the ratios of the sides are only given, being the same with the ratios of the sines of the angles opposite to them.

It is manifest from 47. 1. that of the two sides and hypothenuse, if any two be given, the third may also be found. It is also manifest from 32. 1. that if one of the acute angles of a right angled triangle be given, the other is also given, for it is the complement of the former to a right angle.

If two angles of any triangle he given, the third is also given, being the supplement of the two given angles to two

right angles. Fig. 15. The other cases may be resolved by help of the preceding propositions, as in the following table.

Given. Sought. 1 Two sides, AB, The angles AB:AC :: R:T, B, of AC.

B, C. which Cis the complement. 2 AB, BC, a side The angles BC:BA::R:S, C, of and the hypo- B, C. which B is the complement. thenuse.

R:T,B::BA: AC.

3 AB, B, a side The other
and an angle.side AC.

AB and B, a The hypo- S,C:R::BA: BC.
side and an an- thenuse BC.
gle.

6 BC and B, the The side R:S, B :: BC: CA.
hypothenuse AC,
and an angle.

These five cases are resolved by Prop. 1.

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