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I

VARIETY
Perpendicular Radius.

F the Perpendicular be made Radiv
Bafe becomes Tangent and Hypot
Secant, fee Plate 3. Fig. 3. where it
(from Def. 12 and 14) that as the Pe
cular is made Radius, the Base will be?
and Hypothenufe Secant.

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As Radius 90 1

Is to the Perpend.

So is Secant of the fame

To the Hypothenufe.
And the contrary.

DEMONSTRAT

With the Centre C and any
Radius CD, defcribe an Arch
D E, and erect the Perpendicu-
lar DF which, it is evident, will
be the Tangent, and C F the
Secant of the Arch D E, or
Angle A to the Radius AD.
And in fimilar Triangles CDF,
CBA it will be C D: CB::

A.

DF: BA:: CF; CA. QE D.

F

3

The third Variety Papend "Radius,

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estion 1. In the right angled Triangle C, there are given the Hypothenuse AC ind the Angle A 29° 57', to find the and Perpendicular.

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A

29.57

63

B

LOGARITHMETICALLY. ICA I

The Hypothenufe and Angles are here given to find the Bafe and Perpendicular. Turn to Plate 3, and you will foon perceive (No. 2 and 3) that your Proportion will run thus.

As Secant of the 4C 60° 3' 10'301687
Is to the Hypoth. 63
I'799341
So is Tang. of the ZC 60° 3' 10'239436

To the Bafe 5459

1.737090

And

And for the Perpendicular, it is made Radius, therefore it must be, as fhewn No. 1, Plate 3. But first take either Part of the laft Proportion.

As Secant of the 4C 6030'

Is to the Hypoth. 63

So is Radius 90

To the Perpend. 31°45

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Or, which will anfwer the fame Purpofe.

As Tang. of the 4C 60°1, 301 ́ Î0°239436

Is to the Base 54'59

So is Radius 90

To the Perpend. 31°45

1'737090

10'000000

1*497654

Question 2. In the right angled Triangle A BC, are given the Perpendicular B C 3145 and Angle A 29°57', required the Base and Hypothenufe,

Logarithmetically

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