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3d. By Guntef'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. '

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* :* finding the natural Sines and Co-sines, the reader is referred to Ta€ oa - -

CASE V.

The fiershendicular and * given, to find the angles and - ast.

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In the triangle ABC there is BC 306, and Ac 370 given; to find the angles A and C.; and the base .4B.

1st. By Construction.

Draw a blank line from any point, in which, at B, erect a perpendicular, on which lay BC 306, from a scale of equal parts: from the same scale, with AC 370, in the compasses, cross the first drawn blank line in A, and the triangle ABC is constructed.

Measure the angle A (by prob. 17. sect. 4); and also AB, from the same scale of equal parts the other sides were taken from, and the answers are now found.

The operations by calculation, the square root, Gunter’s scale, and Natural sines, are here omitted, as they have been heretofore fully explained: the statings, or proportions, must also be obvious, from what has already been said.

Answers; The angle A 55° 48'; therefore the angle C 34 12, and AB 208.

CASE VI.

The base and herfiendicular given ; to find the angles and hysiothenuse.

PL. 5. fig. 9.

In the triangle ABC, there is AB 225, and BC 272, given; to find the angles A and C, and the hypothenuse AC.

Ist. By Construction.

Draw a blank line, on which lay AB 225, from a scale of equal parts; at B, erect a perpendicular; on which lay BC, 272, from the same scale : join A and C, and the triangle is constructed.

As before, let the angle A, and the hypothenuse AC be measured; in order to find the anSWerS. -

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By calculation ; the answers from the foregoing proportions are easily obtained, as before.

. But because AC, by either of the said proportions is found by means of a secant; and since there is no line of secants on Gunter's scale; after having found the angles as before, let us suppose AC the radius, and then

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These proportions may be easily resolved, either by calculation, or Gunter’s scale, as before ; and thus the hypothenuse AC may be found without a secant.

From the two given sides, the hypothenuse may be easily obtained, from cor. 1. theo. 14. sect. 4.

Thus the square of AB = 50625
Add the square of BC = 73984

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From what has been said on logarithms, it is plain,

1. That half the logarithm of the sum of the squares of the two sides, will be the logarithm of the hypothenuse. Thus,”

* Demonstration. The square of the hypothenuse of a right-angled triangle is equal to the sum of the squares of the sides, (theo. 14.) hence the Log. of (..AC2 AB2) = the Log, of BC2, and by the nature of Logarithms, the Log. of BC is equal to the Log. of BC2 divided by 2; and in like manner the Log. of (AB2 + BC3) = the Log of MC2, hence dividing the Log. of MC2 by 2 gives the Log. of AC Q. E. D.

The sum of squares, as before, is 124609; its log. is 5.095549, the half of which is 2.547774; and the corresponding number to this, in the tables, will be 353, for AC.

2. And that half of the logarithm of the dif. ference of the squares of AC and AB, or of AC and BC, will be the logarithm of BC, or of AB.

The following examples are inserted for the exercise of the learner. Ex. 1. In the right-angled triangle ABC.

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To find the other two sides. Ex. 4. In the right-angled triangle ABC. the hypothenuse Z. A 62° 40' Given, K.AC 392 poles, the X Ans. {4C 27° 20' (base AB 180 poles BC 348.25 To find the angles and perpendicular.

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