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tan.arc. BC. But DK is the sum of the sines of the arches AB and AC; and KC is the difference of their sines; also BD is the sum of the arches AB and AC, and BC the difference of those arches. Q. E. D.

Therefore, &c.

COR. 1. Because EL is the cosine of AC, and EH of AB, FK is the sum of these cosines, and KB their dif ference; for FK = ¦ FB + EL = EH + EL, and KB = LH-EH-EL. Now, FK: KB:: tan. FDK: tan. BDK; and tan. FDK cotan. DFK, because DFK is the complement of FDK; therefore, FK: KB:: cotan. DFK: tan. BDK; that is FK: KB:: cotan. arc. DB: tan. arc. BC. The sum of the cosines of two arches is therefore to the difference of the same cosines, as the cotangent of half the sum of the arches to the tangent of half their difference.

COR. 2. In the right angled triangle FKD, FK: KD: R: tan. DFK. Now FK = cos AB + cos AC, KD sin. AB+ sin. AC, and tan. DFK-tan. (AB4AC), therefore cos AB + cos. AC: sin. AB + sin. AC::R: tan. (AB+ AC).

In the same manner, by help of the triangle FKC, it may be shown that cos. AB + cos. AC: sin. AC-sin. AB:: R: tan. (AC—AB).

COR. 3. If the two arches AB and AC be together equal to 90°, the tangent of half their sum, that is, of 45°, is equal to the radius. And the arch BC being the excess of DC above DB, or above 90°, the half of the arch BC will be equal to the excess of the half of DC above the half of DB, that is, to the excess of AC above 45°; therefore, when the sum of two arches is 90, the sum of the sines of those arches is to their difference as the radius to the tangent of the difference between either of them and 45°.

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The 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 sides, to the tangent of half their difference.

Let ABC be any plane 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+ sin. C: sin. B-sin. C. But, by the last, sin. B+ sin. C: sin. B-sin. C::

tan.. (B+ C): tan. (BC); therefore also, (11. 5.) CA+AB: CA-AB:: tan. (B+ C): tan. (B—C); Q. E. D.

B

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Let ABC be a triangle; the sum of AB and AC any two sides, 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 meeting BC produced in D, and AC produced in E and F. Join DA, EB, FB; and draw FG parallel to CB, meeting EB in G.

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Because the exterior angle EAB is equal to the two interior 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 CÁD, 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 to 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 difference; and becausè 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. Q. E. D.

PROP. V.

If a perpendicular be drawn from any angle of a triangle to the opposite side, or base; the sum of the segments of the base is to the sum of the other two sides of the triangle as the difference of those sides 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 dif ference of the segments of the base. Q. E. D.

PROP. VI.

In any triangle, twice the rectangle contained by any two sides is to the difference between the sum of the squares of those sides, and the square of the base, as the radius to the cosine of the angle included by the two sides.

Let ABC be any triangle, 2AB.BC is to the difference between AB2+ BC2 and AC2

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A

D

fore also 2BC.BA: 2BC.BD::R: còs B. Now 2BC.BD

is the difference between AB+BC2 and AC2; therefore twice the rectangle AB.BC is to the difference between AB2 + BC2, and AC2 as radius to the cosine of B. Wherefore, &c. Q. E. D.

C

COR. If the radius = 1, BBD=BA x cos B, (1.), and 2BC.BA x cos B=2BC.BD, and therefore when B is acute, 2BC.BA x cos BBC2+ BA2-AC2, and adding AC2 to both; AC2+2 cos B x BC.BA=BC2 + BA2; and taking 2 cos Bx BC.BA from both, AC-BC22 cos B x BC.BA + BA2. Wherefore AC=√(BC2. 2 cos BX BC.BA + BA2).

If B is an obtuse angle, it is shewn in the same way that AC=√ (BC2 +2 cos A × BC.BA + BA2).

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