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ECB, and AEC — ECB=B, the less angle for AC < AB; therefore (II. D. Cor. 2) ECB or CEF=}(C-B). Also, DB = DÀ + AB = AB + AC, and BĚ AB AE = AB - AC. And since angle DCE is a right angle (III. 31), if EC be made radius, DC and FC are tangents of the angles AEC and FEC (Def. 4). But (VI. 2) DB : BE=DČ:CF; or AB+AC:AB- AC=tan ] (C+B): tan (C–B).

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If a perpendicular be drawn from the vertex upon the base of a triangle, the sum of the segments of the base is to the sum of the two sides as the difference between these sides to the difference between the segments of the base.

For (II. c. Cor. 1) the rectangle under the sum and difference of the sides is equal to that under the sum and difference of the segments of the base, therefore (VI. 16) the above proportion subsists.

Schol.-- The preceding propositions are sufficient for the solution of all the cases of trigonometry; but, in particular cases, some of the following propositions may be employed with advantage.

PROPOSITION VIII. In any triangle, twice the rectangle contained by any two of its sides, is to the difference between the sum of their squares and the square of the other side, as radius to the cosine of the angle contained by these two sides.

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

For, from C draw CD perpendicular to AB or AB produced. Then BC:BD=R:cos B. But (VI. 1) 2 AB.BC:

THEOREM.

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2 AB · BD = BC:BD; therefore 2 AB·BC:2 AB· BD =R: cos B. But (II. 12, 13, or Ad. II. 15) 2 AB. BD is

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the difference between ABP + BC2 and AC?; and hence the proposition is proved. Cor.-If the sides opposite to the angles A, B, C, be called a, b, c, respectively, and Radius = 1, then (Qu.

, X. and Ad. V.7) 2ac : a2 + c - 62=l:cos B, or

al + 2 – 62 cos B =

This applies only to the case

2ac in which the angle B is acute. When B is obtuse,

62a2 + c) then cos B =

; but in this case, cos B

2as
ought to be negative, but a? +62-22 is negative, for

c2.
62 = a + c2; hence also in this case, cos B
a2 + c2-22

when its sign is considered. 2ac Taking the side AC instead of AB, which is just changing 6 into c,

a? + 12-02 cos C =

a2

2ab Also, if both sides be multiplied by 2ab, 2ab cos C = al + 12 -- (?, and adding c2 to both, and taking 2ab cos C from them,

c? = a2 + 12 2ab cos C. This applies also to both cases when C is acute and obtuse. Schol.—When a, b, and c, are known, cos C may

be found by the former expression ; and when a, b, and C, are known, ç may be found by the latter.

PROPOSITION IX. In any triangle, the rectangle under the two sides, is to that under the excesses of the semiperimeter above these sides, as the square of the radius, to the square of the sine of half the vertical angle.

Let ABC be any triangle, produce CB till CD=CĂ; join AD, and draw CE

bisecting the angle C, then (I. 4) CE is perpendicular to AD. Draw BF parallel to DE, and BG to CE. Then AC: AE = Rad: sin C

THEOREM.

B

E

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-b).

THEOREM.

(Prop. IV.) if AC be radius; also BC: BF or EG = Rad: sin C; henee (VI. 23, Cor.) AC CB: AE · EG = R2 : sin C. But (II. c. Cor. 1) AE. EG=} (AB + BD)

(AB-BD) in the triangle ABD. But ÀB + BD = ÅB + CD-CB= AB + XC-CB=AB + AC + CB

=–=+ 2CB=2(S-CB), if S = semiperimeter. And AB - BD = AB - (CD-CB) = AB-AC-CB) = )

( AB AC + CB (Ad. II. 2) = AB + AC + CB-2

AC = 2(S-AC); therefore AC.CB:(SAC) (S-BC) =R2 : sina į Č. Cor. If the numerical values of the sides opposite to the

angles A, B, C, be called a, b, c, respectively, and if
8= the semiperimeter and R=1, then

ab:(8-a)(sb) = R2 : sina į C.
or ab. sina į C (s ---a)(3-6), or

(s
sinC=

a) ($

ab PROPOSITION X. In any triangle, the rectangle under its two sides, is to that under the semiperimeter and its excess above the base,

square of the radius, to the square of the cosine of half the vertical angle.

Let ABC be any triangle, and draw the figure ACDH as in the last; produce BF to meet AC produced in E, and draw CG parallel to AD.

Then angle ACH = E, and DCH = CBE, therefore E=CBE, and CE = CB. Also since CG is parallel to AD, the angles at G are right angles. And HG is a rectangle, therefore CH = GF. Now, AČ:CH or GF =R:cos į C, and CE or CB : EG

R:cos E or cos į C; therefore (VI. 23, Cor.) AC CB: EGGF =R?:cos C. But (II. c. Cor. 1, ?

(1 and Ad. II. 15, Sch.) AEG.GF= (AE+AB) (AE-AB) in triangle ABE; and AE + AB =AC + CB + AB=28; and AE- AB = AC + CB - AB=AC + CB + AB-2 AB=2(S-AB), therefore AC.CB:S(S-AB) =R:cos C.

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COR.--Take a, b, c, &c., as in Cor. to last proposition,

and ab : 8 (8 —c) = Ro : cos? C; or if R = 1,
ab.cos?C=(8-c), or

cos' }C='(870).
? =

8 (

ab

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THEOREM.

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PROPOSITION XI. In any triangle, the rectangle under the semiperimeter and its excess above the base, is to the rectangle under its excesses above the two sides, as the square of the radius, to the

square of the tangent of half the vertical angle.

For (Prop. X.) AC-CB:S (S-AB) = R2: cos? {C, and (Prop. IX.) AC CB : (8 — AC) (S-BC) = R2 · S

) : sin? . C, therefore, by direct’equality, S(S-AB):(

SAC) (S-BC) = cos¿C: sinC. But (Cor. 7 to Definitions, and VI. 22, Cor.) cos2 C: sin? C=R2: tana i C, therefore S (

SAB):($ - AC) (S-BC) = RP: tan? į C. COR.-Therefore : (8 -c): (8 - a) (b)= RP: —s s :

8-5 tan? C; and when R = 8 (8 -c) tan? C= (-a) (3-6), or 8

tan” } C = (–a)(8—6)

C

s).

8 (8)

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EXERCISES.

1. If in any triangle a perpendicular be drawn from the Vertex

upon the base, the segments of the base have the same ratio, as the tangents of the parts into which the vertical angle is divided.

2. The base of a triangle is to the sum of its two sides, as the cosine of half the sum of the angles at the base to the cosine of half their difference.

3. The base of a triangle is to the difference of its sides, as the sine of half the sum of the angles at the base to the sino of half their difference.

4. The base of a triangle is to the difference of its segments, as the sine of the vertical angle to the sine of the difference of the angles at the base.

5. Half the perimeter of a triangle is to its excess above

the base, as the cotangent of half either of the angles at the base to the tangent of half the other angle.

6. The excess of half the perimeter of a triangle above the less side is to its excess above the greater, as the tangent of half the greater angle at the base to the tangent of half the less.

7. In a right-angled triangle, radius is to the sine of double one of the acute angles, as the square of half the hypotenuse to the area of the triangle.

8. Radius is to the tangent of half the vertical angle of a triangle, as the rectangle under half the perimeter and its excess above the base to the area of the triangle.

ALGEBRAICAL PRINCIPLES.

I. In Algebra the values of quantities are represented by letters. These values are numbers which denote the number of times that the unit of measure of each kind of quantities is contained in them respectively. These numbers, therefore, may be either terminate or interminate. The algebraical principles employed in the preceding treatise are very simple, and have been already explained, except a few that are referred to in the additional fifth book.

The student may illustrate the propositions in the second book, by taking numbers and products instead of lines and rectangles, and the squares of numbers instead of the

squares of lines. A small ? placed over a number or letter means its square or the product arising from multiplying it by itself. 1. If m =

3, m? = mm, and 32 = 3X3=9. 2. If A be a line 3 inches, A2

square

inches inches 9 square inches, or 9 times the square described on a line of 1 mch. In this case, 1 inch is the unit of measure.

3. Let A = 5 feet, and m a number = 3, then mA = 3 X 5 feet = 15 feet.

4. In the fourth proposition of the second book, let AB=8, AC=5, CB = - 3, then AB2 = AC2 + CB2 +

= 32

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3X3 square

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