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Example 2.—In the triangle a b o, right-angled at b, sup-pose the base, b C, 374| yards, and the angle, A, 52° 8' ;required the other side, a b, and the hypothenuse, a c.

Example 3.—Suppose a ship sail s. w. by w. until shehas made 409 miles of southing; required the distancesailed, and also how far she is west from the meridian of theplace sailed from.

Example 4.—Observing the sun’s altitude to be 30° 45',and the shadow of a tree at the same time to fall 70 feet3 inches distant from the tree, on the horizontal plane;what is the height of the tree, and what will be the lengthof a rope to reach from the extremity of the shadow to thetop of the tree ?

Problem III.— Given the hypothenuse and one side, tofind the angles and the other side.

Example 1.—In the right-angled triangled abc, right-angled at b, let the hypothenuse, a c, be 350 feet, and theperpendicular, a b, 245 feet ; required the angles, A and C,,and the base, b c.

Geometrically.

Draw b C indefinitely toward c ; at the point b makea b = 245, from a scale of equal parts, perpendicular to b c;and from the same scale take a c = 350 ; place one foot ofthe compasses in A, and the other, extending to the base, willcut it in c; then the angles are measured on the line ofchords.

By Calculation.

The hypothenuse A c being radius,

To find angle c.

As ac, 350 . 2.544068

is to ba, 245 . 2.389166

so is radius . 10.000000

to sine of C, 44° 25' 37" .-. 9.845098

To find b C.

As radius. 10.000000

is to cosine of c, 44° 25' 37" . 9.853786

so is A c, 350 . 2.544068

to b c, 249.95 . 2.397854

The perpendicular, a b, being radius,

To find angle a.

As a b, 245 . 2.389166

is to AC, 350 . 2.544068

so is radius . 10.000000

to secant of a, 45° 34' 23" . 10.154902

To find B c.

As radius. 10.000000

is to tangent of a, 45° 34' 23". 10.008688

so is a b, 245 . 2.389166

to bc, 249.95 . 2.397854

The base, bc, being radius,

To find itself.

As tangent of C, 44° 25' 37" . 9.991312

is to radius . 10 000000

so is A b, 245 . 2.389166

to bc, 249.95 . 2.397854

As secant of c, 44° 25'37" . 10.146214

is to radius . 10.00000

so is ac, 350 . 2.544068

to b c, 249.95 . 2.397854

The side bc may also be found, independently of theangles, by means of the known property of a right-angledtriangle ; that the Square of the hypothenuse is equal to thesum of the squares of ths two sides.

For, since ac 2 = ab’ + bc’, it follows, thatBC’ = AC' — ab ! = ( a C + A b) • (a o — a b);and therefore bc=v'(ac + ab)-(ac — ab)

_ _ Log. (ac+ab) + Log. (ac — a jO;

Or, Log. B C =-—- 1 --•_!:--

From which b c is easily determined. j e j

Example 2.—Suppose the hypothenuse of a right' 8,11 gtriangle be 274.5 yards, and its base 196.25 ; what aietwo acute angles ? , gaS t

Example 3.—Suppose a ship sailed between south an ^510 miles, and thereby made her difference of latitu >southing, 315 miles ; upon what course did she sail ?, ^

Example 4.—A ship sailed from latitude 49° 30 n °between the south and west, 135 leagues, till, byobservation, she is found in latitude 45° 15'; requirecourse on which she sailed. r ll( J

Problem IV. — Given the base and perpendicular, t°Jthe angles and hypothenuse. . jjt'

Example 1. —In the right-angled triangle A B C, ® t j l6angled at b, let the perpendicular, ab, be 650 feet, a^ ^base, b c, 420 feet ; required the acute-angles, A and >the hypothenuse, a c.

Geometrically, ^

Make bc = 420, taken from a scale of equal parts > j 0b raise the perpendicular ab = 650; from the° samejoin a c to complete the triangle ; then, with 60° take ^ c ;the line of chords, describe arches round the angles A ft ,jlland their measures, applied to the same line of chor >give the quantity of each angle.

By Calculation,

The base, b c, being radius,

To find angle c. g4 g

Asbc, 420 . 0 8129' 3

is to a b, 650 . in OOOOO 0

so is radius . n t896d 4

to tangent of C, 57° 7' 52" .

rojtni.o. g

As radius. io26 54 ^

is to secant of c, 57° 7' 52". g q 23 %^

so is b c, 420 . g gSSd? 3

to a c, 773.88 .

The perpendicular A b being radius,

To find angle A. gl29' 3

As a b 650 . 2 o 6232*9

is to bc 420.; ioOOOg

so is radius . Q.8W°°

to tangent of a, 32° 52' 8" .

As radius. lO.O? i<?

is to secant of a, 32° 52' 8". 2-3^ a i6

so is a b, 650 . 2-89 3 \i-

to A c, 773.88 ..

The hypothenuse, A c being radius,

T, find iu, V . s.rfg

As sine of c, 57° 7' 52' . \0

is to radius .

so is a b, 650 .'

to a c, 773.88 .