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ABC and AFE being similar, AE: EF:: AC:CB; that is, R: sin. C:: AC: CB (4.6).

2. In any right-angled triangle, radius is to the tangent of one of the acute angles, as the side lying adjacent to this angle is to the side lying opposite.

Draw the perpendicular DG (see the preceding figure), and it will be the tangent of the angle A; then by similar triangles, we have AD: DG:: AB : BC; that is, R: tang. A:: AB : BC.

3. In a right-angled triangle, radius is to one leg, as secant to the hypothenuse. The triangles ADG and ABC are similar; hence AD : AB:: AG:AC; that is, R: AB:: sec. A: hypothenuse.

4. In any plane triangle, the sides are to one another as the sines of their opposite angles.

Let ABC be the proposed triangle; from A let fall the perpendicular AD, which, in the first case, falls within the triangle, and divides it into two right- B angled triangles, viz. ABD and ACD. Then, by Prop. 1, we have

R: sin. B:: AB : AD

R: sin. C:: AC: AD

Hence, sin. C: sin. B:: AB : AC. For,

RxAD=sin. BX AB

R×AD=sin. CX AC

A

C

D

Hence, sin. BxAB=sin. C×AC; then (16.6),

sin. C: sin. B:: AB : AC.

When the extremes are

equal, the means will form a proportion.

A

In the second case, when the perpendicular AD falls without the triangle ABC, in the right-angled triangles ABD and ACD, we have, by D Prop. 1,

R: sin ABD:: AB : AD
R: sin C:: AC : AD

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Hence, sin C: sin ABD :: AB : AC. But as the angle ABD has the same sine as its supplement ABC, we have, by substitution,

Sin C: sin B:: AB: AC.

5. In a triangle, the tangent of half the sum of any two angles is to the tangent of half their difference, as the sum of the opposite sides is to their difference. Let ABC be the proposed triangle, whose angles are A, B, C; and let BC =a, AB=c.

Then

B

A
F

C

D

E

Tan. (A+): tan. ()::a+c:a-c Produce BA, which is not the greater of the two sides under consideration, until DB be equal to BC; join CD, upon which let fall the perpendicular BE, dividing DC into two equal parts. (Cor. 26.1).

The angle BCD=BDC (5.1); therefore BDC= BCA+ACD; but BAC=BDC+ACD (32.1); therefore BAC=BCA+2ACD. From both sides deduct BCA, and we get

2 ACD=BАС-ВСА: hence

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In like manner BAC=BDC+ACD; add BCA to both, and we get BAC+BCA=BDC+ACD+ BCA-2BDC=2BCD; that is,

2BCD=BAC+BCA. Hence we get

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Now AD=BD-AB=BC-AB; and if AD be bisected at F, we have AF-AD BC-AB; and

2

2

2BF-2AB+2AF=BD+AB=BC+AB: therefore BF=BC+AB. By joining EF, it will be parallel to AC, because EF bisects both AD and DC; then (2.6) BE: Eb:: BF: FA. Let the perpendicular BE=y, and bE=x, also EC=z; then from the last analogy we get y : x :: BF : FA, and:::BF: FA:::::a+c:a-c. But CE (z) : EB (y):: R(1): tan. BCD==tan. ; and for the same reason=tan. ACD=tan.; hence tan. : tan.

:: a+c: a-c.

2

2

z

2

In a similar manner it may be shown that tan. B

21

2

2 B-C

2

:

::

tan. :: a+b: a-b; and that tan. B+C : tan. b+c:b-c; which are the general algebraic expressions corresponding to the rule.

6. In a triangle, the cosine of any angle multiplied by twice the product of the sides which contain it, is equal to the sum of the squares of those sides diminished by the square of the third side.

Let ABC be the proposed triangle, and let AB=c,

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