TR I

507

T R I

- j

19 to radius . 10.000000

80 is b c 420 . 2.623249

to Ac, 773.88 . 2.888676

The liypothenuse may also be found, independently of the® n gles, for ac=n/ a b 2 -|- b c 2 ; from which A c is easilyae termined.

Example 2.— In the rectilinear triangle abc, rectangulara * B, suppose the side A B 495.45 yards, and the side B C60.5 yards ; what are the acute angles, A and C ?

Example 3.—Suppose three towns so situated, that a liest miles south from B, and C lies 50^ miles west from B ;le bearing of A from c, and of C from a, are required.Example 4.—When the sun shines, if a steeple, 196 feet"Sh, project a shadow 237 feet 9 inches, on the horizontalP'une, what is the sun’s altitude at that time ?

With respect to taking angles by the theodolite, when wenot obliged to cross zero, subtract the less number ofAgrees from the greater, which will give the angle. But if. e are under that necessity, subtract the greater numberr om 360 and add the less to the remainder, for the angle.

Thus, suppose the index to stand at 145° in looking at oneP°mt or object, and at 253° in looking at another ; or, sup-Po.se first at 253°, and then at 145°, then, in either case,.° 3 °—145° = 108° for the angle contained between the*° objects.

Again: suppose the index stands at 254°, and we areuhged to cross zero in order to come to the other object,then the index to stand at 15° ; or suppose the indexj an ^s at 15° in looking at the first object, and at 254° in°ol<i n g at t) le gecon^ then, in either of these cases, 360° —040 -f 15° = 106° + 15° = 121°.

Eigure 7.— To find the height of an inaccessible object.

4 p et C be the apex of a steeple, standing on the summit ofjj ll3 > and let a b be a straight line parallel to the horizon, so^ at the point c may be seen by a spectator from the points| ' ln< l B, at a convenient distance from the eye above ground :c A B be 367 feet, the angle bac= 44° 30', and the angles * b = 51° 20'; it is now required to find the height of ther j' 1 »it C, of the steeple above the horizon.

, ij V on U

is

subtracting 51° 20' from 180°, a remainder of 128°40'Ij, for the angle ABC; and since the three angles of a^ lle triangle are equal to two right angles, the third anglec c B may be found by subtracting the sum of the two angles

fro B a " d C B A from 180 °> viz ‘ 44 ° 30< + 128 ° 40 ' = 1730 10 'tli e ‘ n . l80 °, "filich will leave a remainder of 6 ° 50'; so that in

r r, ght-angled triangle all the angles and one side are given,

trl! Vh , icl * to find the side bc. Now, as the sides of a plane

Ull gle are as the sines of the opposite angles,

4s

sin e of b c a 6° 50'.

log. =

to

U IS Ato g

sine of b ac, 44° 30' log. =b, 367 feet.log. =

Thus,10 £ 924520

9‘845662

2-564666

which is 2,162 feet, log. = 3‘334848

or thus.9‘075480

9*845662

2*564666

12*410328

9*075480

3*334848

tlle right-angled triangle Arc, right-angled at f, wefof e t le liypothenuse, BC, and the acute angle, c b f; tliere-*i|ies *lie hypothenuse radius, the sides are as the

°t their opposite angles. Wherefore,

As rad. sine of 90° .log. =

is to sine of 51° 20' .log. =

so is 2,162 feet. log. =

to cf, which is 1,688 feet log. =

Thus, or thus.

10 *. = 10*000000

9*892536 = 9*8925363*334848 = 3*334848

3*227384 ... 13*22738410*000000

3*227384

C A =

and in A d b A — =

A d b = 180 -

.". sin

Eigure 8.— To find the distance between two inaccessibleobjects.

Two inaccessible objects, d and c, are both visible fromeach of the two places A and b, whose distance is known,and each of which is visible from the other. Required thedistance c d.

Observe the Z. cab = o, and Z. dab = /3 from A,and Z d b C = a, and Z. CBA=j3 from b.

Let a b = «,

Then, if we can find D a and AC, or D b and b c, theproblem is reduced to this ; find the third side when the othertwo sides and the angle included by them are given.

, T C A sill. C B A sill, /?

Now, in A cab, — 1 = -= -

A B Sill. A C B sin. A C B.

But a c b = 180 (a b c + c a b) = 180 — (/3 + a + /3)

/. sin. a c B = sin. (a -j- jo + (5)a sin. jo

sin. (a + /3 + /3)

da sin. A b D sin. (a -f jo)

a b sin. A d b sin. a d b

(a b d 4 - b A d) == 180 {a + jo + ft) ;adb = sin. Q3 + (3 + a) ;

.". A D — fit . 8i "-< a +/>

sm. (p + p + a)

And thus knowing A C and A D, and Z. C ad, we may, bymeans of a subsidiary angle, compute CD.

TRILATERAL, (from the Latin (res, three, and latus, aside,) a plane figure, or solid angle, having three sides.

TRIM, (from the Saxon trimman, to build,) in general,signifies to fit ; as, to trim up, is to fit up.

TRIMMED : when a piece of work is fitted between twoothers previously executed, it is said to be trimmed inbetween them ; thus, a partition wall is said to be trimnied-up between the floor and the ceiling ; a post between twobeams, a trimmer between joists, &c.

Tkimmed, is also applied to the putting of anything intoshape, by cutting it away by degrees until it be of the pro-posed form.

Tkimmkd-odt, an expression applied to the trimmers ofstairs, when brought forward to receive the rough strings.

TRIMMER, a small beam, into which the ends of severaljoists are framed. Beams of this kind are either stair-trim-mers, hearth-trimmers, or tail-trimmers.

TRIMMING JOISTS, the two joists into which each endof the trimmer is framed. The distance of the trimmijig-joists, when employed in fire-places, must be such as to takein not only tlie fire-place, hut the flues on each side ofit. Trimming-joists ought to be stronger than the otherjoists, on account of the support they have to give.

TRINE DIMENSIONS, the dimensions of a solid,including length, breadth, and thickness; the same as three-fold dimensions.