exceed 45°, are set backwards from 45, and numbered 50, 60, 70, &c.

There is no line of secants on the scale; for every thing requisite can be performed without them.

Thus the two first statings of this case, answer the question without a secant: the like will be also made evident in all the following cases.

CASE II."

The base and angle given; to find the perpendicular and hypothenuse.

fig. 65:

In the triangle ABC there is the angle A 42° 20', and of course of the angle C 47° 40′ (by cor. 2. to theo. 5.) and the leg AB 190, given; to find BC and AC.

Geometrically.

Make the angle CAB (by prob. 16. sect. 1.) in blank lines, as before. From a scale of equal parts, lay 190 from A to B; on the point B, erect a perpendicular BC (by prob. 5. sect. 1.) the point where this cuts the other blank line of the angle, will be C; so is the triangle ABC constructed: let AC and BC be measured from the same scale of equal parts that AB was

Extend from 47° 40', to 90° on the line of sines; that distance will reach from 190 to 257, on the line of numbers, for AC.

2. When AB is made the radius: the first stating is thus performed.

Extend from 45° on the tangents (for the tangent of 45 is equal to the radius, or to the sine of 90° as before) to 42° 20'; that extent will reach from 190, on the line of numbers, to 173, for BC.

3. When BC is made the radius, the second stating is thus performed.

Extend from 47° 40', on the line of tangents, to 45° or radius; that extent will reach from 190 to 173, on the line of numbers, for BC; for the tangent of 47° 40', is more than the radius; therefore the fourth number must be less than the second, as before.

The two first statings of this case, answer the question without a secant.

CASE III.

The angles and perpendicular given; to find the base and hypothenuse.

fig. 66

E

B

In the triangle ABC, there is the angle A 40°, and consequently the angle C 50°, with BC 170, given; to find AC and AB.

As S. A: BC::R: AC,=264.
As R: AC: S. C; AB, 202.

Geometrically.

Make an angle CAB of 40° in blank lines; (by prob. 16. sect. 1.) with BC 170, from a line of equal parts, draw the popped lines EF parallel to AB (by prob. 8. sect. 1.) the lower line of the angle, and from the point

dicular BC (by prob. 7. sect. 1.) and the triangle is constructed the measures of AC and AB, from the same scale that BC was taken, will answer the question.

CASE IV.

The base and hypothenuse given; to find the angles and perpendicular.

In the triangle ABC, there is given, AB 300 and AC 500: the angies A and C, and the perpendicular BC, are required.

As AC: R:: AB: S. C, 36° 52′
As R: AC: S. A : BC,=400. ·

Geometrically.

9

From a scale of equal parts, lay 300 from A to B on B erect an infinite blank perpendicular line, with AC 500, from the same scale, and one foot of the com pass, in A, cross the perpendicular line in C; and the triangle is constructed.

By prob. 17. sect. 1. Measure the angle A, and let BC be measured from the same scale of equal parts that AC and AB were taken from; and you have the

answer.

By Gunter's Scale.

1. Making AC the radius.

Extend from 500 to 300, on the line of numbers ; that extent will reach from 90°, on the line of sines, to 36° 52' for the angle C.

Again, Extend from 90° to 53° 08′, on the line of sines, that extent will reach from 500 to 400, on the line of numbers, for BC.

2. Making AC the radius, the second stating is thus performed.

Extend from radius, or the tangent of 45°, to 53° 08', that extent will reach from 300 to 400, for BC.

CASE V.

The perpendicular and hypothenuse given to find the angles and base.

fig. 68.

In the triangle ABC there is BC 306, and AC 370