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It is also obvious, that AB : AC :: AC: FK, therefore AC2 -=FK, that is to say, the radius is a mean proportional be

AB (tween the tangent and co-tangent of any arch. Hence, the co-tangent may be found by the following rúle

RULE 3. Divide the square of the radius by the tangent of any arch, and the quotient will give the co-tangent of that arch.

EXAMPLE I.

Required the tangent and co-tangent of 60', the co-line being 50000, and sine 86603.

Covine. Sine. Rad. Tang. of 60°.
By Rule 1. 50000 : 86603 :. 100000 : 173206 An

By

Rule 2. to find the cotangent. Sine. Co-fine. 86603 : 50000 :. 100000: 57734 co-tan. of 60°, or tan. of 30

Rad. squared, 10000000000 By ROLE 3

-=57734, as by Rule 2. tan. 60°

173206 Ex. 2. Required the tangent and co-tangent of 40° 30'

.

85409 Ex. 3. Required the tangent and co-tangent of 15° 32'

Anf.

STang 27795

co-tang- 359772 Ex. 4. Required the tangent and co-tangent of 20° 45°

{

Anf. {Tang.

37886

Ico-tang. 263949 Ex. 5. Required the tangent and co-tangent of 80° o'

S 567123

Co-tang. 17632

Anf.

PROBLEM

PROBLEM XXVIII.

To find the secant and co-fecant of any arch, the fine and co-fine bén

ing given.

The figure as in last Prob. CD : CE :: CA : CB; or rather CD : CA :: CA : CB ; therefore the rectangle contained by the co-fine and secant of any arch is equal to the square of the radius : Or the radius is a mean proportional between the co-line and secant of any arch. Hence the secant is found by

RULE I.

Divide the square of the radius by the co-line of any arch; and the quotient will give the fecant of that arch.

Rule 2. Divide the square of the radius by the fine of any arch, the quotient will be the co-fecant of that arch.

EXAMPLE 1.

Required the fecant and co-secant of 60°

By Rule I.
Sq. rad. 10000000000

= 2000co secant of 60° Co-fine of 60°

50000

By Rule II.
Rad. fq. 10000000000

= 115469 co-fecant of 6. Sine of 60°

86603 Secants may also be calculated by 47. I. Euclid, if the radius and tangents are given. Thus, add the squares of the tangent aud radius together, and the square root of their sum will be che fecant.

Ex. 2. Required the secant and co-secant of 24° 13'

Sec. 100640 Anf.

I co-fec. 243789 Ex. 3. Required the secant and co-fecant of 20° 35'

106819

co-fec. 284438 Ex. 4. Required the fecant and co-secant of 10° o'

SSec. 1015424

co. 575871 Ex. 5. Required the secant and co-fecant of 35° 40'

Sec. 123089

Anj. {Sec

PROBLEM. Fig. 84. To find the areas of lunes, or the space included between the inter

secting arches of two circles.

RULE. Find the areas of the two fegments.which form the lune, and their difference will be the area of the lune.

EXAMPLE I. The length of the chord AB is 80, the height DC 20, and DE 8, required the area'of the lune, AEBCA.

AD = 40

AD2=1600
DE 64

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44.72 1

44.721
14.907

59.628
80

139.628

20 vers. Gine.

2792.560

•4
1117.0240 area of ACB.
430.0448 of AEB.

686.9792 area of Lune. Ex. 2. Thę chord is 20, and versed sines 10 and 2. Requited the area of the lune.

Anf. 128.522. Ex. 3. The length of the chord is 48, and the heights of the segments 18 and 7. What is the area? Anf. 405.8676

Note. If semicircles be described on the three sides of a rightangled triangle, as diameters, then will the triangle be equal to the two lunes on the legs, taken together.

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