The perpendicular BC being made radius ; the base AB becomes the tangent of the angle C, or co-tangent of the angle A, and the hypothenuse A C the secant of the angle C, or co-secant of the angle A.—Hence,

To find the Perpendicular BC:

As the angle A 53:7:48" Log. co-secant Ar. compt.

9.903090

As the angle A 53:7:48" Log. co-secant Ar. compt. 9.903090

Is to hypothenuse A C = 246.5 Log. =

2.391817

PROBLEM II.

Given the Angles and One Side, to find the Hypothenuse and the other

Example.

Side.

Let the base AB of the annexed triangle A B C, be 300.5, and the angle A 40:54:40%; required the hypothenuse A C, and the perpendicular B C ?

The hypothenuse A C being made radius; the perpendicular B C will be the sine of the angle A, and the base AB the co-sine of the same angle.

To find the Hypothenuse A C :

As the angle A = 40:54:40% Log. co-sine Ar. compt.

Is to the base A B 300.5

So is radius = 90:

Log.=
Log. sine =

To the hypothenuse AC = 397.6 Log. =

To find the Perpendicular B C :

As the angle A = 40:54:40" Log. co-sine Ar. compt.

Is to the base AB = 300. 5 Log.
So is the angle A

=

40:54:40" Log. sine.

To the perpendicular BC= 260.4 = Log.=.

The base A B being made radius; the perpendicular B C will be the tangent of the angle A, and the hypothenuse A C the secant thereof.-Hence,

2.415647

So is the angle A

To the perpendicular BC= 260.4 = Log. =

The perpendicular B C being made radius; the base A B will be the tangent of the angle C, or co-tangent of the angle A, and the hypothenuse the secant of the angle C, or co-secant of A.-Hence,

To find the Hypothenuse AC:

As the angle A40:54:40" Log. co-tang. Ar. compt.
Is to the base A B = 300. 5 Log. =

So is the angle A= 40:54:40" Log. co-secant =

To the hypothenuse A C = 397.6 = Log. =

9.937802 2.477845 10. 183833

2.599480

To find the Perpendicular BC :

As the angle A

40:54:40 Log. co-tang. Ar. compt. = 9.937802

Given the Hypothenuse and One Side, to find the Angles and the
Other Side.

By making the hypothenuse A C radius; the perpendicular B C becomes the sine of the angle A, and the base A B the co-sine of the same angle.— Hence,

To find the Angle A:

As the hypothenuse A C = 330. 4 Log. Ar. compt. =

Is to radius 90: Log. sine =

So is the base A B 280. 3 Log. =

The base A B being made radius; the perpendicular BC becomes the

tangent of the angle A, and the hypothenuse A C the secant of that angle, -Hence,

Remark. The perpendicular BC may be found independently of the angles by the following rule (deduced from Euclid, Book I. Prop. 47, and Book II. Prop. 5), viz.,

To the log. of the sum of the hypothenuse and given side, add the log. of their difference; then, half the sum of these two logs, will be the log. of the required side :-as thus ;

Given the Base and the Perpendicular, to find the Angles and the

Example.

Hypothenuse.

Let the base AB, of the annexed triangle ABC, be 262. 5, and the perpendicular BC 210.4; required the angles, and the hypothenuse A C?

By making the base A B radius; the perpendicular BC becomes the tangent of the angle A, and the hypothenuse AC the secant thereof. -Hence,

The perpendicular B C being made radius; the base A B will be the tangent of the angle C, or co-tangent of the angle A, and the hypothenuse A C will be the secant of C, or the co-secant of the angle A.-Hence,

To find the Angle A:

As the perpendicular B C 210. 4 Log. Ar. compt. =

Is to radius = 90: Log. sine =
So is the base A B = 262.5 Log. =

To the angle A = 38:42:47" = Log. co-tangent =

To find the Hypothenuse A C :

As radius 90: Log. sine =

=

Is to the perpendicular B C = 210.4 Log.
So is the angle A 38:42:47" Log. co-secant =

To the hypothenuse AC 336.4 = Log. =.

The angle A subtracted from 90: leaves the angle C; thus 90: - 38: 42:47 51:17:13" the measure of the angle C.

=

Remark. The hypothenuse A C may be found independently of the angles by the following rule, deduced principally from Euclid; Book I. Prop. 47; Book II. Prop. 5; and Book VI. Prop. 8, viz.,