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CA S E V.

In the Obl. a dcb, Scd=129

requ.bd

Two Sides, and the Angle comprehended g'ven ; the other side riquired.

Example.

cb=64

, given,

21d=101°15'

Preparation.
Find the Angles bd, by Case the fourth.

Proportion. per. Ax 2, SZb:cd ::S Lc: bd. 5,56°, 15': 139 :: S, 191, 15:164. The Operation is the lame with Case the second.

139 CASE VI,

YA: Fig.62. The three sides given ; the Angles required,

Example.

bd In the Obl. abcd

cd = 70

required the given.

Angles. Preparation by Axiom 4. Froin the Vertical Angle, upon the Base.bd Let fall the Perpendicular Then the { whole, is divided { L'as cad,cab

Segments ad, ab.

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abcd Sbd =105

cb =50

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cd

cd= 70
cb = 50.

Suin = cd + cb = 120.

Diff. =cd-cb= 20.

Proportion per Ax. 4.
db:cd + cb :: cd – cb:df = da-ba
105: 120 : : 20 : 22, 8.

Operation.
To Ar. co. Log. bd = 105 8. 978811

Log. cd -- cb=120. 1:079181 Add

Log.cd-'cb = 20 0.301030 Sum Radius=Log.df 22; 8. 10.359022.

By Gunters Scale.
The extent from IOS
To

I 201 On the Line of Will reach from

20 Numbers.
To

22.8
bd=105=Base.
fd = 22. 8= Differ. Segments,

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Sum 127.8

its 363.9=d=<>Segment: Diff. 82.15

41.7=ba=>S Then the Argles are found by Case 41b of Right Aggled Triangles.

Proportion. In the acab{ fbRad:: ab : 8, 4.0 50:S, 90 ::41.7:9,56. 30.

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od: Rad. :: ad : SLC In the Acad 40: S, 90:: 639:5,465 ° 36".

From 90° OCH
Suh. 56 30.

Remainder - 33,-0 =4b

From 90° COT
Subftract 65 06.

Remainder = 21,54= 4d.

To 56°301
Add 65

06.

Sum=121. 36=lc in a b c d.

Practical Trigonometry. :

Wherein the Do&t. ine of Plane Triarglis are af

plied to Practice.

N this Section, I shall treat only of such

of Plane Right - lined Triangles becomes fu' fervient tc: As,

1. In ALTIMETRIA; By which the Height of any Object accessible, or inacceflible, may be found; As of Trees, Steeples, Towers, &c.

2. In

In LONGIMETRIA: By which the Diftance of one Objet from any place, or of many Objects one from another, whether approachable, or in-approachable may be known, their' Positions laid down, and a Map made of them.

Of Altimetria.
Prob. 1. Of an Altitude that is Acceffible.

Fig. 63. Let AB be a Tower, whose Height you would know. Firfi, At any convenient distance, as at C, place your Quadrant, or any other Inftrument you make use of, and there observe the Angle ACB, which let be 58.°, so much is your Angle of Altitude. Meature next the di. ftance between your Instrument and the Foot of the Tower, viz. The Line CD which let be 25 Yards; then have you in a right Angled Triangle, one Angle c given, and one Leg CB to find the other AB; which you may do as you were taught in Cafe 1. of Trigonometry: For if you take 58 from-90, there remains 32 for the Angle at A. Then say, As clie Sine of the L A 32 9.724210, Is to the Log.of the

1. 39740

Add the two tcBare CD 25

>gether and from So;s the Sine of the LC 58 9.928420 ) the Sum Sub

i stract the first. 11:326360

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To the Loz. Heighe of the

Tower, AB, 40 Yards. S 1.602150. To this 40 Yards, you innit add the heighth ot your Inftruinent from the Ground. In this way of takirg Heights, the Ground ought to be very Level, or you may make great mistakes :

Also the Tower or Tree ihould stand Perpendicular.

Prob. 2. Of an Altitude inacceffible. In the foregoing Figure , let AB be the Tower or Steeple, and suppose CB to be a More, or some other hindrance, that you cannot come nearer then C; plant your Iuftrument, and take the Angle ACB

58 dig. Then go backward any convenient indiftance, as to G, there also take the An

gle AGB 38 deg. This done Sabftract 58 trom 180, lo have you 122 deg, the A gle ACG, then 122 and 38 being taken froin 180, remains 20 for the Angle GC, the distance GC meafured is 26. Trigonometry lay,

As the Sine of the L A 20 9. 534052 is the Lrg.of the diftance GC26 1. 414972 So is the Sine of the Angle G 38 9.789242

I 1.204314 to the Log. of the Line AC 49 1. 670269

Agring

Now by

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