## Trigonometry, Plane and Spherical;: With the Construction and Application of Logarithms |

### Inni boken

Resultat 1-5 av 5

Side 8

it is plain that ADB is

proposed . Moreover , since ABC = ABD ( ADB ) + DBC , and C = ADB - DBC ( by

9. 1. ) it is plain that ABC — C is = 2DBC ; or that DBC is

difference ...

it is plain that ADB is

**equal to half**the fum of the angles opposite to the sidesproposed . Moreover , since ABC = ABD ( ADB ) + DBC , and C = ADB - DBC ( by

9. 1. ) it is plain that ABC — C is = 2DBC ; or that DBC is

**equal to half**thedifference ...

Side 14

Then , the arches BC and CD being equal to each other ( by hypothesis ) , OC is

not only perpendicular to the chord BD ... DG of the two extremes ( because Bm =

Dm ) is therefore

Then , the arches BC and CD being equal to each other ( by hypothesis ) , OC is

not only perpendicular to the chord BD ... DG of the two extremes ( because Bm =

Dm ) is therefore

**equal to half**their fun , and Du**equal to half**their difference . Side 61

and , consequently , half the vertical angle ACB = D + CBD ( by 9. 1. ) = D.

Moreover , seeing DCB is = the sum of the angles A and CBA , at the base ( by 9.

1. ) it is evident that BCF ( or DCF ) is

ECF is ...

and , consequently , half the vertical angle ACB = D + CBD ( by 9. 1. ) = D.

Moreover , seeing DCB is = the sum of the angles A and CBA , at the base ( by 9.

1. ) it is evident that BCF ( or DCF ) is

**equal to half**that sum ; and , therefore , asECF is ...

Side 62

It is B manifest , because CD = CB , that CDB and CBD are equal to one another ,

and that each of them is also

base ( by Cor . 2. to 10. 1. ) ; therefore ABD , being the excess of the greater ...

It is B manifest , because CD = CB , that CDB and CBD are equal to one another ,

and that each of them is also

**equal to half**the sum of the angles CBA and A at thebase ( by Cor . 2. to 10. 1. ) ; therefore ABD , being the excess of the greater ...

Side 65

From whence it is manifeft , thac CD will be

similar criangles AEH and HCD , it will be , AE : CD : :: HA ? = HE X HG ( by Cor ...

From whence it is manifeft , thac CD will be

**equal to half**the sum , and AD**equal****to half**the difference , of the two sides of the triangle . Moreover , because of thesimilar criangles AEH and HCD , it will be , AE : CD : :: HA ? = HE X HG ( by Cor ...

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### Vanlige uttrykk og setninger

added alſo appears arch balf baſe becauſe called caſe chord circle co-f co-fine AC co-tang common complement conſequently Corol COROLLARY determine diameter difference divided drawn equal equal to half evident exceſs extremes fide AC fine fines firſt follows given gives gles greater half the ſum half their difference Hence hyperbolic logarithm hypothenuſe laſt logarithm manifeft meeting method minute moreover Note oppoſite parallel perpendicular plane triangle ABC preceding progreſſion PROP proportion propoſed radius rectangle reſpectively right-angled ſame ſecant ſee ſeries ſhall ſides ſince ſine ſpherical Spherical triangle ABC ſubtracted ſum ſuppoſed tang tangent of half Tbeor Theor THEOREM thereof theſe thoſe unity verſed vertical angle whence