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Example. The height of a tower is 40 feet, and the breadth of its ditch 30 feet : required the length of a ladder to reach from the further side of the ditch to the top of the tower.

1402 + 302 or V1600 + 900, or 2500 50 ft., length of the ladder. COR. II.-If the hypotenuse and one side be given in numbers, the other side may be found: for, let the square of the side be subtracted from that of the hypotenuse, and the remainder is equal to the square of the other side.

The square root of this remainder will therefore be equal to the other side. Thus,

AB

JBC2 – AC?; and AC NBC – AB? Example. One village A is at right angles with two others B and C; the distance from B to C is 50 furlongs ; from A to C 30 furlongs : required the distance from A to B. 1502 – 302 V2500 – 900

V1600 = 40 furlongs from A to B. Cor. III.-If any number of squares be given, a square equal to their sum may be found; or if one square be given, any multiple of it may be ascertained ; or if two squares be given, the difference between them; or a square may be made that shall be the half, fourth, &c., of a given square.

10. Set the st. lines AB and BC, representative of the sides of the first two given squares, G D at rt. angles ABC,—the square on AC A B2 + BC; at. C place CD, representative of the third square, at rt. angles to AC,—the square

AB? BC? + CD2; and at. D place ED, representative of the fourth square, at rt. angles to AD,—the square on AE

A B +

IB BC + CD + DE?.

A And so on for any number of squares.

2o. Supposing AB to be representative of the st. line on which the given square is constructed, its multiple square will be obtained in a similar way; for in this case BC, CD, DE being each equal to AB, the square on A E is the multiple of the square

on AD

on AB.

30. Let A B be the less, and AC the greater of two st. lines; at B the extremity of the less, raise a perpendicular BG, and from A at the other extremity, with AC as radius, inflect on BG the greater st. line; the square of the intercept CB will equal the difference of the squares on AC and AB.

4o Make the angles A and B each equal to

E half a right angle ; C being a rt. angle, the square on AC will be one-half of the square on с.

D AB: again, at. A and. C make the angles CAD, ACD each equal to half a rt. angle, and the

B

A square on CD will be } of that on AC, or of the square on A B : the process may be continued for any bisection of the rt. angles supposed to be formed at the extremities of a line, as for }, 1, 32, &c.

COR. IV.-If a perpendicular B D be drawn from the vertex of a triangle to the base, the difference of the squares of the sides A B and CB, is equal to the difference between the squares of the segments AD and DC. For (47. T.) ABS AD+ BD%;

B

B and CB DC2 + BD? Take the latter from the former,

A B– CBP = ADP – DC?. The difference vanishes when AD DC.

А D CA с

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CoR. V.-If a perpendicular be drawn from the vertex B to the base AC, or AC produced, the sums of the squares of the sides and alternate segments are equal. For A B + BC2 A B + BD+ DC?; and AB? + BC2

BC+ A D2 + BDP; therefore (Ax. 1) A B + BD+ DC2 BC2 + AD2 + BD%. Take away

the common square BD%;
and (Ax. 3) A B + CD2 = BC + A D?.

Sch.—1. The 32nd and 47th Propositions are said to have been discovered by PYTHAGORAS, born B.C. 570, and other principles of Geometry to have been brought by him from Egypt into Greece. Whatever we may think of the tale of his extravagant joy on the discovery of the 47th, certain it is that this is one of the most important propositions in all Euclid, for on it, and on the proposition ir: the sixth book which establishes the similitude of equiangular triangles, the whole science of Trigonometry is founded.

2. A plain and practical illustration of the 47th Proposition may be given by taking three st. lines in the proportion of 3, 4, and 5, and constructing with them a rt, angled triangle BAC : on each of the three sides draw a square, and sub-divide each square ; that on AC 3, into nine smaller squares, that on A B 4, into sixteen, and that on BC 5, into twenty-five squares : the sum of the squares, 9 + 16, on AC and AB, equals the squares on BC, or 25 squares. ,

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To form a rt. angle, st. lines containing 3, 4, and 5 equal parts, or any equimultiples of them, may be used : thus, measure off a st. line containing 5 feet or links, &c.; at one extremity B, with 4 feet or links, draw an arc, and at the other extremity C, with 3 feet or links, another arc : the two arcs intersect in A, and the st. lines from A to C, and from A to B, are at rt. angles to each other.

USE AND APP.-1. The 47th and its corollaries may be applied for the construction of all similar rectilineal figures, by Prop. 31, bk. vi., where it is proved" that in right-angled triangles, the rectilineal figure described upon the side opposite to the right angle, is equal to the similar and similarly described figures upon the sides containing the right angle ;” and to the making of a circle the double or the half of another circle, by Prop. 2, bk. xii., "that circles are to one another as the squares of their diameters."

10. To make a rectilineal figure ADKEF, similar to a given rectilineal figure ABLHC.

E

H

Divide the figure into triangles by the st. lines

F AH, AL produced, if necessary ; take AD equal to the side of the required figure, and through. D draw DK parallel to BL, through.K, KE parallel to LH, C and through. E, EF parallel to HC. Then ADKEF will be similar to ABLHC,-for, by Prop. 29, the fi angles D, K, E, F are equal to those at B, L, H, and C, and the triangles A DK similar to ABL, AKE to ALH, and AEF to AHC.

А

Now, by Prop. 19, bk. vi., Similar triangles are to each other as the squares of their like sides ;and figures made up of similar triangles being similar, all similar figures are to each other as the squares of their like sides. For, area A EF A E2 AK?

AD2

area A EF area AHC area AHC AH?

AL?
A B2
AD2

AB2
area AKE
area ALH
area ADK

area ABL
AD
AB2

AD?

AB?

In like manner

, and

k60°

Adding these equals, we have,
area ADKEF area A BLHC

area ADKEF AD?
AD?

AB?

area A BLHC AB?

-See Tate's Geometry, p. 92. 20. To make a circle the double or the half of another circle. Let AB be the diameter of ADBC; at. A raise a

E perpendicular AE, and at. B make the angle ABE equal to half a rt. angle ; produce | BC until it cuts AE: the square on BE will be double of the square on AB, and the circle of which BE is the diameter double of the circle of which AB is the diameter.

Again, let BE be the diameter of a circle ; at .s E and B make angles each equal to half a rt. angle; and B

A the square on A B will be one-half of the square on BE; and the circle of which AB is the diameter one-half of

D the circle of which BE is the diameter.

2. By this 47th Proposition, the Chords, Natural Sines, Tangents, and Secants of Trigonometrical Tables are constructed.

With st. line AC as radius G desc. an arc CG, and from . A

H raise a perpendicular AG; the

K60° arc CBKG being the measure of a rt. angle, is equal to 90° :

Е. join C and G, CG is the line of I chords ;;--on which the chords of F!

B CB 30°, CK 60°, being inflected, Co is the chord of 30°, Ck the chord of 60°. The sine of arc CB is BD, the tangent CE, the secant A E. The co-sine is FB,

A the co-tangent GH, the co-secant

D AH.

Let it be supposed that the radius AB is divided into 1,000,000 parts, and that the arc BC is 30°. Since the chord Ck of 60° is equal to the radius AC; BD the sine of 30° shall be equal to the half of AC, or 500000, in the rt. angled triangle ADB. Now, AB? = AD? + BD? ; and A B? - BD2 = ADP or BF?, the sine of the complement : substituting the numbers we have v (10000002 - - 5000002) = = 866025 = FB. Next, as the triangles ABD, AEC are equi-angular, we have the proportion

BD. AC, AD : BD :: AC : CE; therefore the tangent of 30° CE

AD 500000 x 1000000 50000000000 or,

= 577350. Then AC? + CE? = A E?, 866025

866025 and A E is the secant of 30° ;—therefore v (10000002 + 5773502) 1154703, the Natural Secant for an arc of 30°.

3. Arithmetically, when the given numbers are 3, 4, 5, or their equimultiples, the sum of the squares of the two less is equal to the squares of the greater, as (6, X 6) + (8 x 8) = 10 x 10 = 100, and when any two are given

WOO

we can find the third exactly : but with respect to all other numbers, though the sum of the squares of any two numbers always equals or constitutes the square of a number greater than either, we cannot attain that number with perfect accuracy ; excepting in the case of right triangular numbers, all we can do is to approach its value by increasing the number of decimal places in the root. Thus (5 x 5) + (8 x 8) = 25 + 64 89; but the square root of 89 is 9.433981, &c. See the Introduction, $ vi., on Incommensurable Quantities. Right triangular numbers may be found thus : n? -1

n2 + 1 Put n any odd number, then the second number, and2

2 49-1

49 + 1 the third number : thus, take 7, then

24, and

= 25.

Now 2

2 (24 x 24) + (7 x 7) = 576 + 49 = 25 x 25 625; the rt. triangular numbers being 7, 24, and 25. 4. The height of any elevation on the earth's

B surface is so small when compared with the earth's diameter, that for practical purposes, as levelling, and ascertaining the height of mountains, we may consider the earth's actual diameter, and the diameter + the elevation, as the same quantity, i.e., BE and L E not sensibly to differ; nor the arc

D AL from the horizontal level AB. We assume LE to be 7960 miles, or that we may bave an easier number, 8000 miles. If we take A B one mile, then BL=gooo part of a mile, or nearly 8 inches ; i.e., for every mile of survey, the surface or curvature of the earth is 8 inches below the horizontal level.

5. Heights and Distances from the curvature of the earth are computed by Prop. 47, from the principle established in Prop. 16, bk. iii., that the tangent A B is perpendicular to the radius CA of the arc A L. Then if AB be required, we have v (LC + L B)2 - A C2 = A B : if L B, the formula is VA B’ + A C2 =BC, and BC-LC or A C=B L.

Example 1. Given B L, the height of the Peak of Teneriffe ; what will be the radius of its horizon, or the distance at which it may be seen ?

Here CB=CL+L B. And VC B? – ACP=A B the horizontal radius.

Or, 40022 – 40002 16016004 - 16000000=16004. And V16004=126 miles =A B, distance at which visible.

Ex. 2. A meteor B is seen over a distance from A to D of 200 miles ; required its height.

Here B L=BC-LC or AC. And VAC? + AB2=C B.
Or, V40002 + 1002 V16010000 4001.24 miles. And 4001.24 - 4000

= 1.24 miles, height of the meteor. Ex. 3. A fountain B one mile from A, is observed from A to have the same apparent level : how much is B above A ? ' i.e., how much is B further from the earth's centre than L or A ?

Here BC-LC = BL. And V (40002 + 12) 4000.0001255 BC: then 4000.001255 - 4000 = .0001255 of a mile = curvature 8 inches nearly.

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