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It is manifest, from Def. 4, that CD is the side of the angle CBF.

Let CB be produced till it meet the circle again in I; and it is also manifest, that AE is the tangent, and BE the secant, of the angle

ABI, or CBF, from Def. 6, 7. Cor. to Def. 4,5,6,7. The sine versed sine, tangent, and secant of an

arch, which is the measure of any given angle ABC, is to the sine, versed sine, tangent and secant, of any other arch which is the measure of the same angle, as the radius of the first arch is to the radius of the second.

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Let AC, MN be measures of the an- B gle ABC, according to Def. 1. ;

OMD ČB the sine, DA the versed sine. AE the tangent, and BE the se. cant of the arch AC, according to Def. 4,5,6,7; NO the side, OM the versed sine, MP the tangent,and BP the secant of the arch MN, according to the same definitions. Since CD, NO, AE, MP are parallel, CĎ: NO : : rad. CB : rad. NB,and AE: MP :: rad. AB: rad. BM, also BE:BP : : AB : BM ; likewise because BC : BD: :BN : BO, that is,B A : BD :: BM : BO, by conversion and alternation, AD: MO :: AB : MB. Hence the corollary is manifest. And therefore, if tables be constructed, exhibiting in numbers the sines, tangents, secants, and versed sines of certain angles to a given radius they will exhibit the ratios of the sines, tangents, &c. of the same

angles to any radius whatsoever. In such tables, which are called Trigonometrical Tables, the radius is

either supposed 1, or some number in the series 10, 100, 1000, &c. The use and construction of these tables are about to be explained.

VIII. The difference between any angle and a right angle, or between any arch and a quadrant, is called


к the Complement of that angle, or of that arch. Thus, if BH be perpendicular to AB, the angle

I CBH is the complement of the angle ABC, and the arch HC the complement of AC; also T the complement of the obtuse angle FBC is the angle HBC, its excess above a right angle; and the complement of the arch I FC is HC.


IX. The sine, tangent, or secant of the complement of any angle is called

the Cosine, Cotangent, or Cosecant of that angle. Thus, let CL or DB, which is equal to CL, be the side of the angle CBH ; HK the tangent, and BK the secant of the same angle; CL or BD is the co

sine, HK the cotangent, and BK tbe cosecant of the angle ABC. Cor. 1. The radius is a mean proportional between the tangent and

the cotangent of any angle ABC ; that is, tad. ABC X cot. ABC


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For, since HK, BA are parallel, the angles HKB, ABC are equal, and

KHB, BAE are right angles ; therefore the triangles BAĖ, KHB

are similar, and therefore AE is to AB, as BH or BA to HK. Cor. 2. The radius is a mean proportional between the cosine and

secant of any angle ABC ; or

cos. ABC X sec. ABC = R'. Since CD, AE are parallel, BD is to BC or BA, as BA to BE.



In a right angled plane triangle, as the hypotenuse to either of the

side, so the radius to the sine of the angle opposite to that side; and as either of the sides is to the other side, so is the radius to the tangent of the angle opposite to that side.

Let ABC be a right angled plane triangle, of which BC is the hypotenuse. From the centre C, with any radius CD, describe the arch DE ; draw DF at right angles to CE, and from E draw ÉG touching the circle in E, and meeting CB in G; DF is the sine, and EG the tangent of the arch DE, or of the angle C.

The two triangles DFC, BAC are equiangular, because the angles, DFC, BAC are right angles, aud

B the angle at C is common. Therefore, CB : BA :: CD: DF ; but CD is the radius, and DF the

D sine of the angle C, (Def. 4.); therefore CB : BA :: R: sin. C.

Also, because EG touches the circle in E, CEG is a right angle,

TE and therefore equal to the angle

Α. BAC ; and since the angle at C is common to the triangles CBA, CGE, these triangles are equiangular, wherefore CA : AB : : CE: EG; but CE is the radius, and EG the tangent of the angle C; therefore, CA: AB : :R: tan. C.

Cor. 1. As the radius to the secant of the angle C, so the side adjacent to that angle to the hypotenuse. For CG is the secant of


the angle G (def. 7.), and the triangles CGE, CBA being equiangular, CA: CB : : CE : CG, that is, CA: CB : : R : sec. C.

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Cor. 2. If the analogies in this proposition, and in the above corollary be arithmetically expressed, making the radius = 1, they give AB AB

sin. C=
; tan. C=


AC'. AC' Also, since sin.C=cos B, because B is the complement of C, cos. B=


and for the same

BC" AC reason, cos. C=


CoR. 3. In every triangle, if a perpendicular be drawn, from any of the angles on the opposite side, the seg- . ments of that side are to one another as

А the tangents of the parts into which the opposite angle is divided by the perpendicular. For, if in the triangle ABC, AD be drawn perpendicular to the base BC, each of the triangles CAD, ABD being right angled, AD: DC : :R: tan. CAD,


D C and AD: DB::R: tap. DAB; therefore, ex æquo,

DC : DB : : tan. CAD : tan. BAD.



The proposition, just demonstrated, is most easily remembered, by stating it thus : If in a right angled triangle the hypotenuse be made the radius, the sides become the sines of the opposite angles ; and if one of the sides be made the radius, the other side becomes the tangent of the opposite angle, and the hypotenuse the secant of it.


The sides of a plane triangle ure to one another as the sines of the oppo

site angles.


From A 'any angle in the triangle ABC, let AD be drawn perpendicular to BC. And because the triangle ABD is right angled at D, AB : AD :: R: sin. B; and for the same reason, AC : AD : : R: sin. C, and inversely, AD : AC : : sin. C:R; therefore, ex æquo inversely, AB : AC: :sin. C: sin. B. In the same manner, it may be demonstrated, that AB : BC :: sin. C :sin. A. Therefore B &c. Q. E. D.




The sum of the sines of any iwo arches of a circle, is to the difference

of their sines, as the tangent of half the sum of the arches to the tangent of half their difference.

Let AB, AC be two arches of a circle ABCD ; let E be the centre, and AEG the diameter which passes through A : sin. AC + sin. AB : sin. AC-sin. AB :: tan. (AC+AB) : tan. Į (AC-AB).

Draw BF parallel to AG, meeting the circle again in F. Draw BH and CL perpendicular to AE, and they will be the sines of the arches AB and AC ; produce CL till it meet the circle again in D ; join DF, FC, DE, EB, EC, DB.

Now, since EL from the centre is perpendicular to CD, it bisects the line CD in Land the arch CAD

c in A : DL is therefore equal to LC, or to the sine of the arch AC; F

B and BH or LK being the sine of AB, DK is the sum of the sines of the arches AC and AB, and CK

Е. is the difference of their sines ;

A DAB also is the sum of the arch

LH es AC and AB, because AD is equal to AC, and BC is their difference. Now, in the triangle DFC, because FK is perpendicular to DC, (3. cor. 1.) DK : KC :: tan. DFK : tan. CFK; but tan. DFK=tan. į arc. BD, because the angle DFK (20. 3.) is the half of DEB, and is therefore measured by half the arch DB. For the same reason, tan. CFK =tan.arc. BC ; and consequently, DK : KC :: tan. arc. BD : tan. į arc. BC. But DK is the sum of the sines of the arches AB and AC; and KC is the difference of the sines ; also BD is the sum of the arches AB and AC, and BC the difference of those arches. Therefore, &c. Q. E. D.


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Cor. 1. Because EL is the cosine of AC, and EH of AB, FK is the sum of these cosines, and KB their difference ; 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. į 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=cog. AB + cos. AC, KD=sin. AB+sin., AC, and

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tan. DFK = tan. 1 (AB + AC), therefore cos. AB + cos. AC : sid. AB + sin. AC :: R: tan. Į (AB + AC). In the same mander, by help of the triangle FKC, it may

be shown that cos. AB+cos. AC : sin. AC-sin, AB :: R : tan, į 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 balf 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° ; there.

; fore, when the sum of two arches is 90°, the sum of the sides of those arches is to their difference as the radius to the tangent of the difference between either of them and 45o.


The sum of any two sides of a triangle is to their difference, as the tan

gent of half the sum of the angles opposite to those sides, to the tangent of half their difference.

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Let ABC be any plane triangle ;
CA+AB : CA--AB :: tan. } (B+C): tan. Į (B-C).
For (2.) CA: AB :: sin. B : sin. C;
and therefore (E. 5.)
CA+AB : CA--AB :: sin. B+sin. C : sin. Bosio. C.
But, by the last, sin. Btsin. C : sin. B-sin. C ::
tan. (B+C): tan. 1 (B-C); therefore also, (11. 5.)
CAFAB CA-AB : : tan} (B+C): tan. } (B-C).
Q. E. D.

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С Otherwise, without the 3d. 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 an

. gles ACB and ABC, to the tangent of half their difference.

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