Again, As Radius

10.000000 Is to the Log. of the Line AC 47 1.672098 So is the Sine of the Angle C 58 9.928420 To the Log. Height of the


11.600518 Tower 40 Yards. B'it as I told you before, the Ground must be Level. However, it it be not, I will


shew you,

Prob. 3. How to take the Hight of a Tower,

&c. when the Ground either vises or falls, Fig. 64 AB is the Tower, CB the Hill whereon you are to take the. Height of the Tower : Plant your Instrument in any place of the Hill, as at C, direct the Sighis to A, and take the Angle A Cd; which let be 19 d g 30 Minutes. Take also the Angle - CB, which is 45, 301; then mead *

. sure the distance CB 56 Yards, take 19° 30% there reinains 70° 30, for the Angle A, then say, As Sine 70° 30

9.974345 is to the distance CB 56 Yards 1.748183

So are both the Angles at C, 9.357276. viz.60°. Oc!.


To the Height ofthe Tower

1. 731118.

54 Yards.

To take this at two Stations, without coming to the Foot of the Tower, is no inore then what has been said before ; for you take your Angles at C, and then mpa. sure to E, and there in the like manner as before, take your Angles again, thereby you inay find all the Angles, and the Line AE, then say,

As the Sine of the Angle_ABE
is to the Log. of the Liue EA,
So is the Sine of the Angle AEB
To the Log. of the H ight of the Tower,


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Pob. 4. How to take the Horizontal-Line of



Fig. 65. Suppose KLMN an Hill, whose Base you would know. Plant

your Instrument at K, and cause a mark to be

up at L, so high above the top of the Hill, as the Instrument ftands from the Ground at K, and take the Argle LKN 58 deg. Measure the distance KL 16 Chains, so L in ks. Then say, 80 As Radius

10,000000 is to the Line KL16 Chains 8oLin. 3.225309 Sois the SineComplement of K.58' 9.724210.


to part of the Base KN 8 Cha. 12.949519

90 Lin.



But if you have occafion to measure the whole Hill, plant again your Instrument at L, (or M) and take your Angle NLM, which let be 46 dig. Measure also the diftance L M 21 Chains: Then fay, As Radius

10 000000 Is to the Line LM 21 Chains 1 222219 So is the Sine of the Angle 9850934

MLN 46 To the part of the Base NM 11.179153.

is Chains 12 Lin. Which 15 Chains 12 Lin, added to 8 Chains 90 Lin. makes 24 Chains, 2 Lin. for the whole Base KM.

I mentioned this way, for to make you understand how to take part of a Hill; for many times your Survey may end on the Side of a Hill. But if you find you are to take in the

! whole Hill, you need not take so much pains as the foriner way. But thus: Take, as before, the Angle K 58 deg. Measure KL. Then at L take the Angle KLM 78 deg. SukHract those 2 from 180 deg. Remains 44 for the Angle at M, Then say,

As she Sine of the Angle M.
is to the Log. of the Side KL;
So is the Sine of the Angle at L
to the Log. of the Base KM.

Prob. 5.

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How to take the Altitude of an Ob. je&t At.mnding upon a Hill, inaccefible. Fig. 66. Suppose NO to be the Object, and you standing at P were requis red to find the heighth thereof.

First, upon Paper draw a right Line at pleasure, as QT, and make choice of any Point at pleasure, as at P, for the place of your fianding; then with a Quadrant directed to the Top of the Object, you find the Degrees cut to be 40°.52', and then direct the Sights to the Bottom of the Object at 0, and let the Degreees cut be 22.25

Then upon P, protract an Löf 40°52' and draw the Line P w at pleasure : And an 6 of 22°. 25', and draw the Lire Pc at pleasure,

Secondly, go forwards in a right Line to wards the Object as at R, 212 5 feet; and there direct your Sights to the top of the Object at N, and you will find the degrees cut to be 61° E2', through which draw a line at pleasure, as RS, crossing the line P w in the point M, the top of the Olyject: From whence à Perpendicular let fall vpon the Ground-line P T, as NK; that Line shall be equal to the Altitude and of the Hill, and the Object together, ( and direc the Sights to 0, and drum the Line Rt, cutting the Line Pc in the point o.) D 2




Now by the Intersections of these four Lines, Pw, Rs, Pc, Rr, there is constituted tour Triangles, viz. PNK, and RNK, bo h Right Angled at K, and PNR, and RNO Oblique angled: By the resolving of which from the distance measured PR, and the fevera: Angles observed, at R and P, you may find the required Altitude,

I. In the Oblique angled Triangle PNR, there is given, the Angle NPR 40-52', and the Angle NRP 118° 28'; for it is the Corrplement of the Angle NRK 61*821 to 180°. And the Side measured PR, 212.5 Foot, And having the Angles at R and P, the Sum of them is 159o. 10', which take froin 180, there will remain 20° so', for the Angle PNR. From which Triangle given, the two olher Sides PN, and NR may be found by Axioin 2. thus,

As Sine of the L PNR 20°50.
Is to the Side RP Log. 2725
Si is the Sine of NPR 40° 52
To the Side NR

320 And so is the Sine of the Complement of 61.82' = 118° 28' (or the 2 NRP) To the Side PN, 515, 66.

In the Right Angled. Triangle RNK there is given, (1) The Hypothenuse RN= 980 Foot (2) The Angle' NRK 61.3:2 d. Whereby you may fiud NK, by Cafe 1, of Right Angle plain Triangles; thus


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