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THE 'HE sum of any two sides of a triangle is to their

difference, as the tangent of half the sum of the angles opposite to those fides, to the tangent of half their difference.

Let ABC be any plain triangle ; CA+AB:CA-AB:;tan. (B+C):tan. (B-C). For (2.) CA: AB : ; sin. B : sin. C; and therefore (E. 5.) CA + AB:CA--AB :: sin. B+fin. C: fin. B - fin. C. But, by the last, fin. B+fin. C; fin. B - fin. C:: tan. (B+C): tan (B-C); therefore also, (11. 5.) CA + AB:CA — AB::tan. Ź (B+C) : tan Ġ (B-C), l. E. D.

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Otherwise, without the 3d.

Let ABC be a triangle ; the sum of AB and AC any two fides, is to the difference of AB and AC, as the tangent of half the sum of the angles ACB and ABC, to the tangent of, half their difference.

About the centre A with the radius AB, the greater of the two sides, describe a circle meet.

G ing BC produced in

B
D, and AC produ-
ced in E and F.
Join DA, EB, FB;
and draw FG paral.
lel to CB, meeting
EB in G.

IC
E

F
Because the exte.

А rior angle EAB is equal to the two intcrior ABC, ACB, (32. 1.); and the angle EFB, at the circumference is equal to half the angle EAB at the centre (20. 3.); therefore EFB is half the sum of the angles opposite to the sides AB and AC.

Again, the exterior angle ACB is equal to the two interior CAD, ADC, and therefore CAD is the difference of the angles ACB, ADC, that is of ACB, ABC, for ABC is equal to ADC. Wherefore also DBF, which is the half of CAD, or BFG, which is equal to DBF, is half the difference of the angles opposite to the sides AB, AC.

Now because the angle FBE in a semicircle is a right angle, BE is the tangent of the angle EFB, and BG the tangent of the angle BFG to the radius FB; and BE is therefore tó BG as the tangent of half the sum of the angles ACB, ABC to the tangent of half their difference. Also CE is the sum of the sides of the triangle ABC, and CF

their

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their difference; and because BC is parallel to FG, CE: CF:: BE:BG, (2. 6.); that is, the sum of the two sides of the triangle ABC is to their difference as the tangent of half the sum of the angles opposite to those sides to the tangent of half their difference. . E. D.

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F a perpendicular be drawn from any angle of a

the segments of the base is to the sum of the other two sides of the triangle as the difference of those fides to the difference of the segments of the base.

For (K. 6.), the rectangle under the sum and difference of the segments of the base is equal to the rectangle under the sum and difference of the sides, and therefore (16.6.) the sum of the segments of the base is to the sum of the sides as the difference of the sides to the difference of the segments of the base. Q. E. D.

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N any triangle, twice the rectangle contained by

any two fides is to the difference between the sum of the squares of those fides, and the square of the base, as the radius to the cofine of the angle included by the two sides.

Let

D

Let ABC be any triangle, 2 AB.BC is to the difference be. tween AB+BC? and AC? as radius to cos B.

From A draw AD perpendicular to BC, and (12. and 13. 2.) the difference between the sum of the squares of AB and BC, and the square on AC is equal to

B 2BC.BD.

But BC.BA: BC.BD::BA : BD::R: cof B, therefore also 2BC.BA : 2BC.BD::R: cos B. Now 2BC.BD is the difference between AB? + BC and AC?, therefore twice the rectangle AB.BC is to the differeace between AB? + BC”, and AC2 as radius to the cofine of B. Wherefore, &c. Q. E. B D.

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Cor. If the radius = 1, BD = BA X cof B, (1.), and 2BC.BA x cos B=2BC.BD, and therefore when B is acute, 2BC.BA X cof B=BC? + BA? - AC?, and adding AC? to both; AC? + 2 cof B X BC.BA=BC? + BA?; and taking 2 cos BXBC.BA from both, ACBC2- -2 cof BXBC.BA+ BA?. Wherefore AC=N(BC2—2 cof B X BC.BA+BA?.).

If B is an 'obtufe angle, it is shewn in the same way that AC=N (BC? + 2 cof B x BC.BA + BA?.).

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F

YOUR times the rectangle contained by any twg

sides of a triangle, is to the rectangle contained by two straight lines, of which one is the base or third side of the triangle increased by the difference of the two fides, and the other the base diminished by the difference of the same, fides, as the square of the radius to the square of the fine of half the angle included between the two sides of the triangle.

Let ABC be a triangle, of which BC is the base, and AB the greater of the two lides : 4AB.AC: (BC+(AB-AC)) (BC — (AB -- AC)): R2 : (sin BAC)?.

Produce the side AC to D, so that AD = AB; join BD, and draw AE,CF at right angles to it; from the centre C with the radius CD describe the femi.

H circle DGH, cutting BD in K, BC in G, and meeting

D BC produced in H.

F It is plain that B CD is the difference of the sides, and therefore that BH is the base increased, and BC the base diminished by the difference of the sides ; it is also evident, because the triangle BAD is isosgeles, that DE is the half of BD, and DF is the half of DK, wherefore DE – DF=the half of BD- DK, (6. 5.) that is EF=BK. And because AE is drawn parallel to CF, a fide of the triangle CFD, AC : AD: : EF : ED, (2. 6.); and rectangles of the same altitude being as their bases AC.AD: AD2: : EF.ED: ED?, (:. 6.), and therefore 4AC.AD: AD? : : 4EF.ED: ED?, gr alternately, 4AC.AD : 4EK.ED :: ADP : ED?

But

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