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The solutions of the several problems in heights and distances, by Gunter's scale, are omitted ; because every particular stating has been already shewn by it, in the rectangular and oblique angular trigonometry:

OF DISTANCES.

Any of the instruments used in surveying, will give you the angles or bearings of lines; which will be particularly shewn, when we come to treat of them.

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Let A and B be two houses on one side of a river, whose distance asunder is 293 perches : there is a tower at C on the other side of the river, that makes an angle at A, with the line AB of 53° 20'; and another at B, with the line BA of 66° 20' : required the distance of the tower from each house, viz. AC and BC.

This is performed by case 2. of oblique angled trigonometry, thus :

1. As S.C : AB :: S. A: BC,=270.5

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Let B and C, be two houses whose direct distance asunder, BC, is inaccessible; however it is known that a house at A, is 252 perches from B, and 230 from C; and that the angle BAC, is 70°.

What is the distance BC, between the two houses ?

This is performed by case 3. of oblique angular trigonometry, thus;

1. As AB+AC: AB-AC:: T. of 1 C+B:

T of C-{B=3° 44'.

55° +3° 44'=58° 44'=0.550—3°-44=51° 16'=B.

2. AS S.C: AB :: S. A: BC,=277.

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Suppose ABC a triangular piece of ground, whichi by an old survey we find to be thus : AB 260, AC 160, BC 150 perches; the mearing lines, AC and BC, are destroyed, or plowed down, and the line AB, only remaining. What angles must be set off at A and B, to run new mearings by, exactly where the old ones were ?

This is performed by case 4. of oblique angled trigonometry, thus;

1. As AB : AC+BC: : AC-BC: AD-DB=11. 92; then,

130+5.96=135.96=AD.
130_-5.96=124.04=DB.

2. As AD:R:: AC : Sec. A,=31° 47'

3. As BC.: S. A:: AC: S. B,=34° 10'.

PROB, IV.

fig. 90.

D

B

Let D and C, be two trees in a bog, to which you can have no nearer access than at A and B; there is

49°, and the line AB 113 perches. Required the distances of the trees DO.

180°—the sum of DBA and DAB=ADB=31o. 180° the sum of CAB and CBA=ACB=22.30.

In the triangle ABD, find DB, thus ;
1. As S. ADB: AB :: S. DAB : DB,=216.

And in the triangle ABC, find BC thus ;

2. As S. ACB : AB :: S. CAB : BC,=175.6

In the triangle DBC, you have DBC=ABC-ABD =72°; likewise the sides BD, BC, as before found, given to find DC.

3. As BD+BC: BD—BC:: T. of į DCB+{CDB: T. of į DCB-CDB,=8° 5'

54° +8° 05=62° 05=DCB.
54°-9° 05=45° 55=CDB.

4. As S. CDB : BC :: S. DBC : DC,=232.5

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