I wished to know the distance between a kirk and a mill, which were upon the other fide of a river, I choose two ftations, 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.

In the triangle AKM, to find the angles AMK, MKA.

13.58350 the lefs

2137

2

To tan. AMK-MKA 48° 23′10. 05138

Note, The foregoing example may be performed, by ufing MB and BK as the containing fides.

EXAMPLE VIII. Plate 5. fig. 6.

If the Peak of Teneriff be four miles above the level of the fea, and the angle of depression taken from the farthest visible point, be 87° 25′ 55′′. Required the diameter of the earth, alfo the fartheft vifible point that can be feen from the Peak.

If the fquare 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 fea, the quotient will give the earth's diameter, and the height of the spectator's eye above the level more.

Demon. Because the ftraight line AC is equally divided at E, and produced to the point D, the rectangle AD, DC, together with the fquare of EC, is equal to the fquare of ED, but the fquare of ED is equal to the fquares EB, BD, because DBE is a right angle; therefore, the rectangle AD, DC, together

with the fquare of EC-EB, is equal to the fquares EB, BD; take away the common fquare EB, and the remaining rectangle AD, DC, is equal to the fquare of BD the visual ray. And because the rectangle AD, DC, is equal to the fquare of BD, (Euclid. 17th. 6.) DC: DB::DB: AD.: Therefore, DB AD and AD-DC-CA the diameter.

Here it must be observed, that if from any point without a circle, two ftraight 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 178.452=7961

4

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

Several methods have been invented to find the earth's diameter. Mr Picart of the Academy of sciences at Paris, has propofed 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 toifes, each of which con tains 6 feet of the fame 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 femidiameter of the earth, 19.615,800 feet: but the French ma'thematicians, who, of late, examined Mr Picarts observations, • affure us, that a degree in that latitude, is 57.183 toifes. They measured a degree in Lapland, in the latitude of 66° 20′, • and found it to be 57.438 toises. By comparing these degrees, as well as by the obfervations on pendulums, and the theory ' of gravity, it appears, that the earth is an oblate spheriod; and the axis or diameter that paffes 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 confiderably 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 denfity 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.'

PROBLEM III. Plate 5. fig. 9.

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

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

venient diftance 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.

EXAMPLE.

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

5: 4 :: 80

4

5)320

65=AB.

PROBLEM IV.

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

Place any staff of a known length in the fame 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.

EXAMPLE.

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 4 feet, and of the fhadow, 6 feet, required the height of the steeple.