DOI Page / Citation link:
https://doi.org/10.11588/diglit.20658#0024
10
OF DRAWING.
Case 2. When the given degrees are more than 60.—After having described the arch as before,
take one half or one third of the given number of degrees, and lay the transverse distance of this
one half or one third off, upon the described arch, twice if the degrees were halved, and three
times if the third part were taken.
Case 3. When the required angle is less than six degrees.—Example : It is required to con-
struct an angle of 4 degrees : first, construct an angle of 60 degrees, as by the first case, then set
off the chord of 56 degrees from the same angle, on the arch thereof; and the remaining part
of the arch will be the arch of the required angle of four degrees.
To find the length of the sine, tangent, and secant, of a given angle, having a given radius.
Suppose the given radius be 3 inches, and the given angle 27 degrees SO minutes ; open the sector
till the transverse distance of 90, and 90 on the line of sines, be equal to the given radius (3
inches); then the transverse distance of 27 degrees and an half on the line of sines will be the
length of the sine for that angle and radius; and the transverse distance of 27 and an half on the
tangent line will be the length of the tangent for the same angle and radius ; and for the length
of the secant, make the given radius (3 inches) the transverse distance from 0 to 0 at the begin-
ning of the secant line ; and the transverse distance of 27 and one half on the same line, will be
the length of the secant required.
If the tangent of an arc of more than 45 degrees be required, the transverse distance of 45 at
the beginning of the scale of upper tangents is to be made equal to the given radius ; and the
length of the required tangent is to be taken from this scale in a transverse distance, at the given
number of degrees.
The scales of upper tangents and secants do not extend to 76 degrees ; therefore, if the tangent
or secant of an arc of more than 76 degrees be required, they will be found by making the radius
of the circle a transverse distance to the complement, from 90 of those degrees on the lower tan-
gent line, for the tangent required : and for the secant, making the radius the transverse distance
to the complement of the sines: but in this case the radius of the circle must not be too large,
otherwise it cannot be made the transverse distance.
To find the length of the radius from having the length of the sine, tangent, or secant, of any de-
gree. Make the given length a transverse distance to its given number of degrees, on its respective
scale : then, the transverse distance of 90, and 90 on the sines ; or the transverse distance of 45,
and 45 on the lower tangent line; or the transverse distance of 0, and 0 on the secant line, will
be the radius sought.
And having the radius and the length of the sine, tangent, or secant, to find the number of degrees
corresponding to such sine, tangent, or secant. Set the sector to the given radius, according as
the sine, tangent, or secant is concerned : take the length of the sine, tangent, or secant between
the compasses: apply the two feet transversely to the scale of sines, tangents, or secants; and
where both points rest on the same division on either of these lines, there will the number shew
the degrees and parts of the angle corresponding to the sine, tangent, or secant.
'To find the length of a versed sine to a given number of degrees, and a given radius. Make the
transverse distance of 90 and 90 on the sines, equal to the given radius : take the transverse distance
of the sine complement of the given number of degrees: then if the given number of degrees be
less than 90, the difference between the sine complement and the radius gives the versed sine; but
if the given degrees exceed 90, the sum of the sine complement and radius is the versed sine.
The
OF DRAWING.
Case 2. When the given degrees are more than 60.—After having described the arch as before,
take one half or one third of the given number of degrees, and lay the transverse distance of this
one half or one third off, upon the described arch, twice if the degrees were halved, and three
times if the third part were taken.
Case 3. When the required angle is less than six degrees.—Example : It is required to con-
struct an angle of 4 degrees : first, construct an angle of 60 degrees, as by the first case, then set
off the chord of 56 degrees from the same angle, on the arch thereof; and the remaining part
of the arch will be the arch of the required angle of four degrees.
To find the length of the sine, tangent, and secant, of a given angle, having a given radius.
Suppose the given radius be 3 inches, and the given angle 27 degrees SO minutes ; open the sector
till the transverse distance of 90, and 90 on the line of sines, be equal to the given radius (3
inches); then the transverse distance of 27 degrees and an half on the line of sines will be the
length of the sine for that angle and radius; and the transverse distance of 27 and an half on the
tangent line will be the length of the tangent for the same angle and radius ; and for the length
of the secant, make the given radius (3 inches) the transverse distance from 0 to 0 at the begin-
ning of the secant line ; and the transverse distance of 27 and one half on the same line, will be
the length of the secant required.
If the tangent of an arc of more than 45 degrees be required, the transverse distance of 45 at
the beginning of the scale of upper tangents is to be made equal to the given radius ; and the
length of the required tangent is to be taken from this scale in a transverse distance, at the given
number of degrees.
The scales of upper tangents and secants do not extend to 76 degrees ; therefore, if the tangent
or secant of an arc of more than 76 degrees be required, they will be found by making the radius
of the circle a transverse distance to the complement, from 90 of those degrees on the lower tan-
gent line, for the tangent required : and for the secant, making the radius the transverse distance
to the complement of the sines: but in this case the radius of the circle must not be too large,
otherwise it cannot be made the transverse distance.
To find the length of the radius from having the length of the sine, tangent, or secant, of any de-
gree. Make the given length a transverse distance to its given number of degrees, on its respective
scale : then, the transverse distance of 90, and 90 on the sines ; or the transverse distance of 45,
and 45 on the lower tangent line; or the transverse distance of 0, and 0 on the secant line, will
be the radius sought.
And having the radius and the length of the sine, tangent, or secant, to find the number of degrees
corresponding to such sine, tangent, or secant. Set the sector to the given radius, according as
the sine, tangent, or secant is concerned : take the length of the sine, tangent, or secant between
the compasses: apply the two feet transversely to the scale of sines, tangents, or secants; and
where both points rest on the same division on either of these lines, there will the number shew
the degrees and parts of the angle corresponding to the sine, tangent, or secant.
'To find the length of a versed sine to a given number of degrees, and a given radius. Make the
transverse distance of 90 and 90 on the sines, equal to the given radius : take the transverse distance
of the sine complement of the given number of degrees: then if the given number of degrees be
less than 90, the difference between the sine complement and the radius gives the versed sine; but
if the given degrees exceed 90, the sum of the sine complement and radius is the versed sine.
The