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12.41356 To the distance 266

2.42484 EXAMPLE VII. Plate 5. fig. 5: I wished to know the distance between a kirk and a mill, which were upon the other side of a river, I choose two stations, A and B, distant 400 links, and found the angles MAK 40°, KAB 64° 25', and ABM 56° 15', MBK 50° 8'. Required the distance between K the Kirk, and M the Mill.

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170 48

180 00

9 12 ang. AKB.

In the triangle AKM, to find the angles AMK, MKA. As AK+AM 3405

3.53212 to } fum 70° oo is to AK---AM 1395

3.14457 | add dif. 48° 23' So is tan. AMK+MKA 70° 10 43893

the greater 118 23

13.58350 | the less To tan. AMK-MKA 48° 23'10. 05138


21 37


To find the distance between M and K. As fine angle MKA 21° 37' 9:56631 is to MA 1905

3.00217 So is fine angle MAK 40°. 9.80807

12.81024 To the dist. of the objects 1754 3.24393

Note, The foregoing example may be performed, by ufing MB

and BK as the containing sides.

EXAMPLE VIII. Plate 5. fig. 6.

If the Peak of Teneriff be four miles above the level of the sta, and the angle of depression taken from the farthest visible point, be 87° 25' 55". Required the diameter of the earth, also the farthest visible point that can be seen from the Peak.

If the square of the visual ray, being a tangent to the earth, be divided by the height of the spectator's eye, above the level of the sea, the quotient will give the earth's diameter, and the height of the spectator's eye above the level more.

Demon. Because the straight line AC is equally divided at E, and produced to the point D, the rectangle AD, DC, together with the square of LC, is equal to the square of ED, but the square of ED is equal to the squares EB, BD, because DBE is 2 right angle; therefore, the rectangle AD, DC, together

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with the square of EC=EB, is equal to the squares EB, BD;

away the common square EB, and the remaining rectangle AD, DC, is equal to the square of BD the visual ray. And because the rectangle AD, DC, is equal to the square of BD, (Euclid. 17th. 6.) DC : DB : : DB : AD.: Therefore, DBP =AD and AD-DC=CA the diameter.


To find FO.

To find CF. As rad. 90° 10.000oO As rad. 90°

I0.00000 is to DC 4 0.60206 is to DC 4.

0.60206 So sec. 87° 25' 55" 11.34866 So is tan. 87° 25' 55" 11.34822

TO FD 89.27

1.95072 | To CF 89.18


Here it must be observed, that if from any point without a circle, two straight lines be drawn to touch the circle, they are equal to one another, (Eucl 37. 3.); therefore, FC is equal to FB, but BF and FD make up BD the visual ray; consequently, it will be 89.18+89.27=178.45=BD, and 178.45=7961


=AD, and 7961–4=7957, the earth's diameter nearly.

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Several methods have been invented to find the earth's diameter, Mr Picart of the Academy of sciences at Paris, has proposed an exact method, by which, not only the equatorial and polar diameters may be known, but also the figure of the earth determined.

According to Mr Picart, ' a degree of the meridian at the latitude of 49° 21', was 57.06 French toises, each of which con. ( tains 6 feet of the same measure; from which it follows, that <if the earth be an exact sphere, the circumference of a great

circle of it, will be 123.249,600 Paris feet, and the semi• diameter of the earth, 19.615,800 feet: but the French mathematicians, who, of late, examined Mr Picarts observations,

affure us, that a degree in that latitude, is 57.183 toises. * They measured a degree in Lapland, in the latitude of 66° 20', 6 and found it to be 57.438 toises. By comparing these degrees, as well as by the observations on pendulums, and the theory

of gravity, it appears, that the earth is an oblate spheriod; 6 and the axis or diameter that passes through the poles, will be to the diameter of the equator, as 177 is to 178, or the earth will be 22 miles higher at the equator, than at the poles. A

degree has likewise been measured at the equator, and found
• to be considerably less than in the latitude of Paris, which
' confirms the oblate figure of the earth. Hence it appears,

that if the earth were of an uniform density from the surface
to the centre, then according to the theory of gravity, the me-
ridian would be elliptical, and the equatorial would exceed
the polar diameter, by about 44 miles. '

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PROBLEM JII. Plate 5. fig. 9.

To find the height of an object, by means of one fiaff:

Suppose the pole AB of an unkown height, BC a horizontal plane, and ED a staff of a known length. At any conve



venient distance from the pole, fix your staff perpendicular in the ground, then move backwards or forwards, till you find the point C, whence you may view the top of your staff, E, in a line with A the top of the object, then say, as CD:DE:: CB : BA the height of the object Fig. 67. plate 5.


Let BC be 80 feet, CD 5, and DE 4, required AB.

5:4: : 80





To measure the height of an object from the length of its fbadow.

Place any staff of a known length in the same plane with the object; then say, as the length of the staff's shadow, is to the length of the staff; fo is the length of the object's shadow: to its height.


Wanting to know the height of a steeple, whose shadow I found to be 200 feet, I fixed my staff perpendicular to the horizontal plane, the length of the staff, is 41 feet, and of the thadow, 6 feet, required the height of the steeple.

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