Or, using the logarithms, it will be log. A B+log sin A+log sin B+log.cosec A CB+log sec D BE - 40 (in the index) = log CD; in the case proposed=log. of 83.9983 feet.

EXAMPLE VII.

Α

In order to determine the distance between two inaccessible objects E and w on a horizontal plane, we measured a convenient base AB of 536 yards, and at the extremities A and B took the following angles, viz. B A w=40° 16', w A E=57° 40', A B E=42° 22', E B W=71° 7'. Required the distance E w.

First, in the triangle A B E are given all the E angles, and the side A в to find B E. So, again, in the triangle A B w, are all given the angles, and A B to find в W. Lastly, in the triangle BE W are given the two sides E B, B w, and the included angle E B w to find E w=939:52 yards.

B

W

Remark. In like manner the distances taken two and two, between any number of remote objects posited round a convenient station line, may be ascertained.

EXAMPLE VIII.

b

B

Suppose that in carrying on an extensive survey, the distance. between two spires A and B has been found equal to 6594 yards, and that C and D are two eminences conveniently situated for extending the triangles, but not ad- A mitting of the determination of their distance by actual admeasurement: to ascertain it, therefore, we took at c and d the following angles, viz. SAC B=85° 46' SAD C=31° 48'

BC D=23° 56'

D

A D B=68° 2' Required C D from these data.

In order to solve this problem, construct a similar quadrilateral A c d b, assuming c d equal to 1, 10, or any other convenient number: compute Ab from the given angles, according to the method of the preceding example. Then, since the quadrilaterals A c d b, A C D B, are similar, it will be, as a b c d :: A B C D ; and c d is found = 4694 yards.

EXAMPLE IX.

Given the angles of elevation of any distant object, taken at three places in a horizontal right line, which does not pass through the point directly below the object; and the respective distances between the stations; to find the height of the object, and its distance from either station.

E

B

Let A EC be the horizontal plane, F E the perpendicular height of the object above that plane, A, B, C, the three places of observation, F A E, F BE, F C E, the angles of elevation, and A B, B C, the given distances. Then, since the triangles A E F, BEF, CEF, are all right angled at E, the distances A E, B E, C E, will manifestly be as the cotangents of the angles of elevation at A, B, and c.

Put A B=D, B c=d, E F = x, and then express algebraically the theorem given in Geom. Triangles, 25, which in this case becomes,

A E. BC + CE3 . AB = BE2.AC+AC.AB.BC.

The resulting equation is

dx cot A+D x cot c=(D+d) x cot3 B+ (D+d) D d.
From which is readily found

(D+d) Dd

x= d cot A+D cot c-(D+d) cot B

Thus EF becoming known, the distances A E, BE, CE, are found, by multiplying the cotangents of A, B, and c, respectively, by E F.

Remark.-When D=d, or D+d=2 D=2 d, that is, when the point B is midway between A and c, the algebraic expression becomes,

x=d÷(cot A+ cot c— - cot B),

which is tolerably well suited for logarithmic computation. The rule may, in that case, be thus expressed.

Double the log. cotangents of the angles of elevation of the extreme stations, find the natural numbers answering thereto, and take half their sum; from which subtract the natural number answering to twice the log. cotangent of the middle angle of elevation: then half the log. of this remainder subtracted from the log. of the measure distanced between the first and second, or the second and third stations, will be the log. of the height of the object.

The distance from either station will be found as above.

Note. The case explained in this example, is one that is highly useful, and of frequent occurrence. An analogous one is when the angles of elevation of a remote object are taken from the three angles of a triangle on a horizontal plane, the 'sides of that triangle being known, or measurable but the above admits of a simpler computation, and may usually be employed.

From a convenient station P, where could be seen three objects, A, B, and c, whose distances from each other were known (viz. A_B = 800, A c = 600, B C = 400 yards), I took the horizontal angles A P ̊C 33° 45', BP c = 22° 30'. It is hence required to determine the respective distances of my station from each object.

=

Here it will be necessary; as preparatory to the computation, to describe the manner of

ва

from

any con

C

D

B

Construction.-Draw the given triangle A B C venient scale. From the point a draw a line A D. to make with A B an angle equal to 22° 30', and from в a line в D to make an angle DBA 33° 45'. Let a circle be described to pass through their intersection D, and through the points A and B. Through c and D draw a right line to meet the circle again in P so shall be the point required. For, drawing P A, P B, the angle APD is evidently A B D, since it stands on the same arc A D and for a like reason B P D = B A D. So that is the point where the angles have the assigned value.

=

The result of a careful construction of this kind, upon a good sized scale, will give the values of P A, P C, P B, true to within the 200dth part of each,

Manner of Computation.—In the triangle A B C, where the sides are known, find the angles. In the triangle A B D, where all the angles are known, and the sides A B, find one of the other sides A D. Take B A D from в A C, the remainder, D A C is the angle included between two known sides, A D, A C ; which the angles A D C and A C D may be found, by chap. iii. case 2. The angles CAP = 180° (A PC+AC D). Also, BCP = BCA — A CD: and P B C A B C + P BA = A B C +

from

sup. A D C. Hence, the three required distances are found by these proportions. As sin A PC: AC:: sin PAC: PC, and :: sin

PCA: PA; and lastly, as sin в PC: BC:: sin B P C: B P. The results of the computation are, P A = 709.33, P c = 1042·66, PB 934 yards.

**The computation of problems of this kind, however, may be a little shortened by means of an analytical investigation. Those who wish to pursue this department of trigonometry may consult the treatises by Bonnycastle, Gregory, and Woodhouse.

Note. If c had been nearer to P than A B, the general principles of construction and computation would be the same; and the modification in the process very obvious.

II. Determination of Heights and Distances by approximate Mechanical Methods.

1. For Heights.

1. By shadows, when the sun shines.-Set up vertically a staff of known length, and measure the length of its shadow upon a horizontal or other plane; measure also the length of the shadow of the object whose height is required. Then it will be, as the length of the shadow of the staff, is to the length of the staff itself; so is the length of the shadow of the object, to the object's height.

R

2. By two rods or staves set up vertically.-Let two staves, one, say, of 6 feet, the other of 4 feet long, be placed upon horizontal circular or square feet, on which each may stand steadily. Let A B be the object, as a tower or steeple, whose altitude is required, and A c the horizontal plane passing through its base. CD and E F, the two rods, be placed with their bases in one and the same line c A, passing through a the foot of the object; and let them be moved nearer to, or farther from, each other, until object is seen, in the same line as D and F,

Let

D

CE

G

H

the summit в of the of the rods. the tops

Then by the principle of similar triangles, it will be, as D H (CE): FH:: DG (= C ̧A): BG; to which add A G for the whole height A. B.

= CD,

3. By Reflection.-Place a vessel of water upon the ground, and recede from it, until you see the top of the object reflected from the smooth surface of the liquid. Then, since by a principle in optics, the angles of incidence and reflection are equal, it will be as your distance measured horizontally from the point at which the reflection is made, is to the height of your eye above the reflecting surface; so is the horizontal distance of the foot of the object from the vessel to its altitude above the said

surface.*

4. By means of a portable barometer and thermometer.— Observe the altitude B of the mercurial column, in inches, tenths, and hundredths, at the bottom of the hill, or other object whose altitude is required; observe, also, the altitude, b, of the mercurial column at the top of the object; observe the temperatures on Fahrenheit's thermometer, at the times of the two barometrical observations, and take the mean between them. B - b

Then 55000 X B+b= height of the hill, in feet, for the tem

perature of 55° on Fahrenheit. Add of this result for every degree which the mean temperature exceeds 55°; subtract as much for every degree below 55°.

This will be a good approximation when the height of the hill is below 2000; and it is easily remembered, because 55°, the assumed temperature, agree with 55, the effective figures in the coefficient; while the effective figures in the denominator of the correcting fraction are two fours.

*Where great accuracy is required logarithmic rules become necessary, of which various are exhibited in treatises on Pneumatics. The following, by the Rev. W. Galbraith, of Edinburgh, is a very excellent approximation.

+ h (0.00268 +0.00268 cos 2 + 0.00000005 h)

in which н is the true height in feet, t the temperature of the

* Leonard Digges, in his curious work the Pantometria, published in 1571, first proposed a method for the determination of altitudes by means of a geometrical square and plummet, which has been described by various later authors, as Ozanam, Donn, Hutton, &c. But as it does not seem preferable to the methods above given, I have not repeated it here.