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The second case is solved by means of the third and fourth propositions of this book. Thus, if a, b, C be given, take C from 1800, and (I. 32.) the remainder is A + B. Take the half of this, and then, by the third proposition, as a +b:a-b:: tan}(A + B):tan (A— B). This analogy gives half the difference of A and B ; and* by adding this and }(A + B) together, A, the greater angle, is obtained, while B is found by taking }(A - B) from $(A + B). The remaining side will be calculated (App. III. 4.) by means of either of the following analogies; and by employing both, an easy verification of the process is obtained :

-as cos}(A—B): cos}(A + B):: a +b:c; and sin}(A,B): sin}(A + B)::a-bic.

When the given angle C is a right angle, the solution is most easily effected by means of the first proposition of this book; the oblique angles being obtained by the analogy, a : 6::R : tanB, or cotA ; and the hypotenuse either by the analogy, R : secB : : a : c, or sin A :R:: :c.t

The THIRD CASE may be solved by means of the fifth, sixth, or seventh proposition. Thus, assuming a (see the figures for the fifth proposition,) as base, we have a to b + c, as b to a fourth proportional. If this be less than BC, it is the difference of the segments BD, DC, in the first figure ; and if half of it and half of the base be added together, the sum will be the greater segment, while the less will be found by taking half that proportional from half the base. If the fourth proportional be greater than the base, it is the sum of the segments in the second figure, and, as before, the segments are the sum and difference of half the proportional and half the base. Then, by resolving, by the first case, the two right-angled triangles ADB, ADC, in which there are given the hypotenuses AB, AC, and the legs BD, CD, the angles B, and ACD will be obtained, which, in the first figure, are two of the required angles; while in the second, the angle C is the supplement of ACD.

Again, by adding the three sides together, and taking half the sum, the value of s is obtained : and, in applying the sixth pro

c, or c-6

* See the note to the 13th proposition of the second book.

† As an example, let a= 57.38 miles, b= 42:6 miles, and C= 56° 45'; to resolve the triangle.

:6-8

Asa +6

99 98 1.999913) Ascos (A-B) 15° 18'9-981311 | Assin?(A-B) 9:421662 14.78

1169674 : cos(A+B) 98676913 : sin (A + B) 9-944411 : : tan}(A + B) 61° 37'} 10 267498 : :a+b

I'999913

1:169674 : tan!(A - B)

15° 18'}
9137259
49-26 1.692515

49.26 1 *692459 Hence A = 76° 56', and B= 46° 19'

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Half the difference of A and B is here taken as 15° 18'1. When determined accurately, however, it is found to be 15° 18' 23". Hence the cause of the slight difference in the logarithm of c, as obtained by the two different analogies. It is plain, that after A and B are computed, c might also be found by means of the first case, by either of the analogies ; sin A : sin C : : a : c, and sio B : sin C :: b:c. The foregoing method, however, is much preferable.

position, the sides containing the required angle are to be separately taken from s; but in applying the seventh, only the side opposite to the required angle is to be subtracted; wbile if all the three sides be subtracted successively another method of solution is furnished by the second and third corollaries to the seventh proposition. This last method is preferable to any of the others, when it is necessary to determine all the angles; and if they be all computed by means of it, the correctness of the operation is ascertained by trying whether their sum is 1800.*

Schol. 2. The following problems exhibit some easy and useful applications of plane trigonometry.

PROB. I.

Let it be required to find the height of an accessible object AB, standing on a horizontal plane.

On the horizontal plane take a station C, and measure with a Jine, a chain, or any such instrument, the distance CB to the base of the object; and with a quadrant, a theodolite, or other angular instrument, measure the angle BCA, called the angle of elevation. Then, since B is a right angle, the height AB will be found (APP. III. 1.) by the analogy, R: tan C::CB: BA.

This gives the height of A above CB, the horizontal line passing through the eye of the observer: and therefore to find the

B

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subt. tani A

* To exemplify the last of these methods, let a=679, b= 537, and c=429; to compute the angles.

Here, by adding the three sides together, we obtain 1645, the half of which, 822:5, is s. Tben, by taking from this the three sides successively, we find s-a

143.5, 8-b= 285.5, and 8-c= 8.5. The rest of the operation, the subtraction in the first part of which may be performed in the manner pointed out in the example for the first case, is as follows: 8225 2-9151367

9.9892812
add.

12.1461332
1135 2-1568525
log (s-a) 2:156852

subt.

log (sc) 2 594915 $ 2855 245566151 393 5 2591915

12:1461337

subt.

tan C 19° 35' 9-551188 log (5-6) 2 455606

C=39° 10 2) 19-978563

tan B 26° 73 96905 tand A 44° 17'} 9-989281

B= 520 15 A = 890 357

In the first part of this operation, the halving of the logarithm serves for the extraction of the square root. The remainder of the work consists in adding together tant A and log(s-a), and subtracting log(sb), and log(s—c) successively from the sum. This method of solution is remarkably easy, requiring for the entire operation only four logarithms to be taken from the tables ; and affording at the same time a most satisfactory verification by the addition of the three angles, when found. The preparatory part of the process also admits of an easy verification, as the sum of the three remainders s--a, s--b, 8-c is equal to the balf sum.

entire height, AB must be increased by the height of his eye above the base of the object. The like addition must be made in every problem of this kind, when the angle of elevation above the horizontal line is given,

PROB. II.

To find the height of an object AB, standing on a horizonal plane, but inaccessible on account of the unevenness of the ground near its base, or the intervention of obstacles.

In a straight line passing through the base of the object take two stations C, D; and measure CD, and the two angles of elevation BCA, BDA. Then (I. 32.) CAD is the difference of ACB, ADB, and (App. III. 2.) sinCAD : sin D :: DC: CA. Again, (App. III. 1.) R: sin ACB :: AC: AB; whence AB will be found.

The computation will be rendered rather more easy by multiplying together (V. Sup. 15.) the terms of the two analogies, and dividing the third and fourth terms of the result by AC; as by this means we get the analogy R X sinCAD : sin D x sin BCA :: DC : AB. Hence, to find the logarithm of AB, to the logarithm of DC add the logarithmic sines of D and BCA, and from the sum take the sine of CAD and the radius.

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PROB. III.

To find the distance of two objects A and B, on a horizontal plane.

This may be effected in different ways according to circumstances.

1. A base AC may be taken, terminated at one of the objects. The angles A and C, and the side AC are then measured; and the required distance AB is found by the analogy, sin B : sinC :: AC: AB.

2. This method fails, if the objects A and B be not visible from one another, as then the angle A cannot be measured. In this case, a station C may be taken as before, from which both A and B are visible. Then, having measured the angle C, and the sides AC, BC, we compute the distance AB by means of the second case of trigonometry.

3. When, from inequalities in the ground, or other causes, the

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preceding methods are inapplicable, the solution may be effected in the following manner.

Measure a base CD, such that A and B are both visible from each of its extremities: measure also the two angles at C, and the two at D. Then, by the first case in trigonometry, we compute AC in the triangle ACD, and BC in the triangle BCD: from which, and from the contained angle ACB, AB is computed by means of the second case. The operation may be verified by computing AD, BD, by the first case, and thence AB by the second case.

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D

E

D

PROB. IV. LET AFB be a great circle of the earth, supposed to be a sphere; E a point in the diameter BA produced, EF a straight line touching the circle in F, and ED a straight line in its plane perpendicular to AB: it is required to compute the angle DEF, and the straight line EF.

Draw the radius CF. Then, since (III. 18.) CFE is a right angle, we have (hyp. and I. 32. cor. 3.) DEC = CEF + ECF. Take away CEF, and there remains DEF = ECF. Now (III. 36.) EF = BE.EA. Hence EF will be found by adding AE to AB, multiplying the sum by EX, and extracting the square root.

To find CE, add AE to the radius AC. Then (App. III. 1.) CE: EF :: R: sin ECF, or sin DEF; or CE: CF::R: cos ECF, or cos DEF. Schol. The foregoing solution gives correct results in every

When AE, however, is inconsiderable in comparison of AB, a very near approximation to the true results is obtained, in an easy manner, by using BA and CA instead of BE and CE, and the arc AF instead of EF. In this way, putting n equal to the feet in AE, since a mile is equivalent to 5280 feet, we have EF=V (BA.AE) = V(7912 x 5280 X n); or, in miles, EF = V(1912 X 5280 x'n)= | 79358"

3n, very nearly, since 5280 the sum of 5280 and its half is 7920, which differs

very

little from 7912. Hence, therefore, to find EF in miles, to AE in feet, and its half, and extract the square root.

Again, (APP. I. 40. schol.) the circumference of AFB = 7912 Xt; and as 7912 XT: AF, or EF:: 3600 : ACF, or DEF: that is,

7912 x n as 7912 X7 :

:1791 :: 3600: DEF. Hence, by dividing the

5280

case.

n

DEF =

first and second terms by 7912, reducing the third to seconds, and dividing the product of the means by the first term, weget, in seconds,

360 X 3600

Х

7912

X 5280
360 X 3600
XV 2n = 901 XIV

2n,
V(2 x 7912

x 5280) very nearly; as is easily shown by substituting (App. I. 40. schol.) for # its value 3.141593, and computing the multiplier of V 2n, which is readily found to be 45], or 901 x }, nearly. To find DEF, therefore, in seconds, double AE in feet, extract the square root, and multiply by 901.

If E be the position of the eye, the angle DEF is called the dip of the horizon, and F is a point in the line in which the sky and sea appear to meet, and which is called the offing. *

E

D

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B

*Hence, if AE, the height of the eye, be six feet, the distance of the offing will be three miles ; if 24 feet, 6 miles ; if 32 feet, 7 miles, nearly; if 600 feet, 30 miles ; if 3000 feet, 67 miles, nearly, &c. If, again, the height of the eye be 8 feet, the dip will be 180", or 3', nearly; if 18 feet, 4' 31", nearly; if 1000 feet, 33' 38", nearly, &c.

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