To find how high a person at Philadelphia, must be raised into the atmosphere, and how far the sight must eatend, to see London; the distance being 51 Degrees, on the Rotundity of the Earth, and the Diameter of the Earth 7964. JMiles.


From right to left, draw a line, to represent the earth’s diameter, on which describe a circle with the chord of 60° ; from the centre, raise a perpendicular, and where it cuts the circle, will represent Philadelphia, as at A, from which lay off 51 degrees, to represent London, as at B; from the centre, draw a line to B, and on the Ppoint B, raise a perpendicular to intersect the former at V ; then,

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in the right angled triangle CBV, the angle VCB 51°, and base CB 3982 miles, are given to find the hypothenuse CW, from which, take the earth’s semidiameter, leaves AV 2345 miles, the height required; and VB,4917 miles, is the distance the sight must extend, to see London.

Mote. The operations the same as Case 1, Right Angled Trigonometry.

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LET A and B be two houses on one side of a river, 293 perches asunder ; and a tower at C, on the opposite side of the river, which makes an angle at A, with the line AB cf 53° 20'; and at B, with the line BA of 66° 20'. Required, the distance from the tower to each house?

Jims To A 308.8 perches e { To B 270.5 perches

JMöte. This is performed by Case 1,
Qblique Angled Trigonometry.



Let B and C be two houses, and an observer at A 252 perches from B, and 230 from C, finds that they make an angle at A of 70°. The distance between the houses is required 2 o

Answer, 277 perches.


JVote. This is performed by Case 4 and 5, Oblique
Angled Trigonometry.


Let D and C be two trees in a bog, and an observer at A and B 113 perches asunder, finds, that when at A, the tree at D makes an angle with the line AB of 100°; and that at C, makes an angle with the line AB, of 36° 30' ;

and when at B, the tree at D makes an angle with the line BA of 49°, and that at C of 121°.

Required the distance these trees are asunder?

Answer 232.5 perches.

Mote. By Case 1, Oblique Angled Trigonometry, find the distance from

cither station to each tree, as from B to D 216 perches; and to C 175,6 perches; then, by Case 4 and 5, find their distance apart.

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Multiply the base into the perpendicular, and if they be in chains, divide the product by 10, for acres; and multiply the remainder by 4, for roods, and by 40 for perches; But if in perches, divide by 40 and by 4, for acres, roods and perches, as in the subsequent example.


Let ABCDA be a square field, and each side 7 chains 29 links. Required the content, and to lay down a map by a scale of ten perches to an inch.

Chains. Perches. . . .

7.29 - 29.16 7.29 =: 29.15 A. 5,3,1441 4,0)85,0.3056 4. *- 3)21 10 P. R. 1,257.64 ... -

40 A. 5 1 10.3056

P. 10,30560

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