## The Elements of Euclid: Also the Book of Euclid's Data ...cor. viz. The first six books, together with the eleventh and twelfth |

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Resultat 1-5 av 5

Side 3

other arch which is the measure of the fame angle , as the

the

according to Def . 1. CD the fine , DA the versed fine , AE the tangent , and BE the

secant ...

other arch which is the measure of the fame angle , as the

**radius**of the first is tothe

**radius**of the second . Let AC , MN be measures of the angles ABC ,according to Def . 1. CD the fine , DA the versed fine , AE the tangent , and BE the

secant ...

Side 4

5 : PROP . I. Na right angled plain triangle , if the hypothenuse be made

the sides become the fines of the angles opposite to them ; and if either side be

made

...

5 : PROP . I. Na right angled plain triangle , if the hypothenuse be made

**radius**,the sides become the fines of the angles opposite to them ; and if either side be

made

**radius**, the remaining fide is the tangent of the angle opposite to it , and the...

Side 22

therefore AF is the tangent of the arch AC ; and in the rectilineal triangle AEF ,

having a right angle at A , AE will be to the

angle AEF , ( 1. Pl , Tr .; ) but AE is the fine of the arch AB , and AF the tangent of

the ...

therefore AF is the tangent of the arch AC ; and in the rectilineal triangle AEF ,

having a right angle at A , AE will be to the

**radius**as AF to the tangent of theangle AEF , ( 1. Pl , Tr .; ) but AE is the fine of the arch AB , and AF the tangent of

the ...

Side 24

of the angle ECF opposite to it , that is , in the triangle ABC , the co - fine of the

hypothenuse BC is to the

tangent of the angle ACB . Q. E. D. Cor . 1. Of these three , viz . the hypothenuse

and ...

of the angle ECF opposite to it , that is , in the triangle ABC , the co - fine of the

hypothenuse BC is to the

**radius**, as the co - tangent of the angle ABC is to thetangent of the angle ACB . Q. E. D. Cor . 1. Of these three , viz . the hypothenuse

and ...

Side 27

The rectangle contained by the

the rectangle contained by the co - lines of the opposite parts . These rules are

demonstrated in the following manner . First , Let either of the sides , as BA , be ...

The rectangle contained by the

**radius**, and the fine of the middle part is equal tothe rectangle contained by the co - lines of the opposite parts . These rules are

demonstrated in the following manner . First , Let either of the sides , as BA , be ...

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added alſo altitude angle ABC angle BAC baſe becauſe Book Book XI caſe circle circle ABCD circumference common cone cylinder Definition demonſtrated deſcribed diameter divided double draw drawn equal equal angles equiangular equimultiples exceſs fame fides firſt folid fore four fourth given angle given in poſition given in ſpecies given magnitude given ratio given ſtraight line greater Greek half join leſs likewiſe magnitude manner meet muſt oppoſite parallel parallelogram perpendicular plane priſm produced PROP proportionals Propoſition pyramid radius reaſon rectangle rectangle contained remaining right angles ſame ſecond ſegment ſhall ſhewn ſides ſimilar ſolid ſphere ſquare ſquare of AC taken THEOR theſe third thro triangle ABC wherefore whole

### Populære avsnitt

Side 157 - D ; wherefore the remaining angle at C is equal to the remaining angle at F ; Therefore the triangle ABC is equiangular to the triangle DEF. Next, let each of the angles at C, F be not less than a right angle ; the triangle ABC is also, in this case, equiangular to the triangle DEF.

Side 24 - ... be equal, each to each ; and also the third angle of the one to the third angle of the other. Let ABC, DEF be two triangles which have the angles ABC, ECA equal to the angles DEF, EFD, viz.

Side 45 - If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square of the aforesaid part.

Side 73 - THE straight line drawn at right angles to the diameter of a circle, from the extremity of...

Side 163 - If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.

Side 81 - ELF. Join BC, EF ; and because the circles ABC, DEF are equal, the straight lines drawn from their centres are equal: therefore the two sides BG, GC, are equal to the two EH, HF ; and the angle at G is equal to the angle at H ; therefore the base BC is equal (4.

Side 323 - Equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides.

Side 80 - EA : and because AD is equal to DC, and DE common to the triangles ADE, CDE, the two sides AD, DE are equal to the two CD, DE, each to each ; and the angle ADE is equal to the angle CDE, for each of them is a right angle ; therefore the base AE is equal (4.

Side 36 - To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.