C A S E V.

}

requ.bd.

Two Sides, and the Angle comprehended gʻven ; the other side r quired.

Example
In the Obl. a dcb, Scd=129

cb=64
given,

21d=101°15'

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

Proportion. per. Ax 2. S,Lb:cd :: Slc: bd. 9,56°, 15":139::S, 191, 15 :164. The Operation is the lame with Case the second.

tis

640 CAS E VI.

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

Example. In the Obl. abcd Sbed

required the given.

Angles. Preparation by Axiom 4. Froin the Vertical Angle, upon the Base.bd Let fall the Perpendicular

wholea, 7 is divided { L'As cad, cab Then the { whole is divided { Fale 3

Segments ad, ab.

cd

bd =105 cd =70 cb =50

-ca,

cd= 70
cb = 50.

=

Sumn - od + 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 1:079181Add Log.cd - cb = co 0.301030 Sum_ Radius=log.df 22; 8. 10.359022.

By Gunters Scale. The extent from

105 To

I 2 On the line of Will reach from

20 Numbers.
To

22.8
bd=105=Base.
fd = 22.8= Differ. Sgments,

Sum 127.8

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

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

Proportion.
cb Rad. . S
50:.

In the scab{56 : 5,90 :: 41.7:5,56. 30.

L.c

(od: Rad. :: ad : SLC In the acad

70: S, 90:: 639:5,465° 369. From 90° OCH Suh, 56 30,

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

Practical Trigonometry.

Wherein the Do&t. ine of Pline Triargl

. s are af plied to Practice.

1 N this Section, I shall treat only of such

Practical parts thereof, as the Doctrine of Plane Right - lined Triangles becomes fulfirvient rc: As,

1. In ALTIMETRIA; By which the Height of any Object acceslible, inacceflible, may be found; As of Trees, Steeples, Towers, Etc.

2. In LONGIMETRIA : By which the Diftance of one Obj er 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 thein.

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

Fig. 63. Let AB be a Tower, whose Height you would know. Firfl, At any convenient distance, as at C, place your Quadrant, or any other Instrument you make use of, and there observe the Angle ACB, which let be 58.0, 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-go, there remains 32 for the Angle at A. Then say, As clie Sine of the L A 32 9.724210, Isto the Log. of the

1, 39740

Add the rwo to Bare CD 25

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

stract the first. ¥1.326360

1

To the Log. Height of the ?

Tower, AB, 40 Yards. $1.602150. To this 40 Yards, you innst add the heighth ot you: Inftruinent from the Ground. In this way of taking Heights, the Ground ought to be very Level, or you may make great mistakes i also the Tower or Tree Thould tand Perpendicular.

Prob. 2. Of an Altitude inacceffible.
In the foregoing Figure, let A B
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 I:ftrument, and take the Angle ACB
58 dig. Then go backward any convenient
distance, 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.

Now by
Trigonometry lay,
As the Sine of the L A 20 9. 534052
is the Lng.of the distance GC26 1. 414972
So is the Sine of the Angle G 38 9.789282

I 1.204314 to the Log. of the Line AC 47 1. 670269