Or having found one of the required sides, the other may be obtained, by one or the other of the cors. to theo. 14. sect. 1.

By Gunter's Scale.

1. When AC is made the radius.

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 tangent (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 172 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 perpendiculars given; to find the base and hypothenuse.

Plate V. Fig. 6.

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.

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 where it cuts the other line in C, let fall a perpendicular 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.

What has been said in the two foregoing cases, is sufficient to render the operations in this, both by calculation and Gunter's scale, so obvious, that it is needless to insert them; however, for the sake of the learner, we give for

Answer, AC 264.5, and AB 202.6.

CASE IV.

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

Plate V. fig. 7.

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

Geometrically.

From a scale of equal parts, lay 300 from A to B; on Berect an infinite blank perpendicular line, with AC 500, from the same scale, and one foot of the compass, 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 cor. 2, theo. 5. 90°-36° 52′=53° 08′ the

Or BC may be found from cor. 2. theo. 14.

sect. 1.

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 52°. 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.

N