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SOLUTION OF RIGHT ANGLED TRIANGLES, WITH THE AID

OF LOGARITHMS.

EXAMPLE.

61. Referring to the example of Art. 39, where the hypothe

nuse

1010 × 15

a

sin 35c

employing 1010 as R, instead of 1, because the tables which we are about to use, are constructed with that value of R, we have, by the rules for multiplication and division of logarithms

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Subtract log. sin 35° = 9.758591

Remainder 1.417500=log. of 26.15 the hyp.

30°

EXAMPLE II.

200 feet.

Referring to Art. 41 where the height of the tower c is

ctan 30° × 200

which becomes, when the radius 1 in the first term of the proportion is changed into 1010

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62. It is customary, where the subtraction of logarithms corresponding to the division of numbers is to be performed, to change this operation into addition by means of what is called the arithmetical complement of each subtractive logarithm. The arithmetical complement of a logarithm is the remainder after taking the logarithm from 10; thus the arithmetical complement of the logarithm 2.322447, is 7.677553.

10.000000

2.322447

7.677553

It may be formed most conveniently, instead of beginning at the right and subtracting each figure from 10 and carrying one each time throughout, by beginning at the left and subtracting each figure from 9 till you come to the last figure which subtract from 10.

When we have a logarithm to subtract, we shall obtain the same result by adding its arithmetical complement and afterwards subtracting 10. Which may be proved as follows: By the definition arith. comp. log. b 10 log. b. Now add this arith. comp. to some other log. as log. a, the result will be

=

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Hence to perform operations containing a number of multiplications and divisions, by means of logarithms, we have the following

RULE.

Write the logarithms of all the multipliers, and the arith. complements of those of the divisors in a column. Add up the whole, and reject as many times 10 from the characteristic of the sum as there have been arith. complements employed.

Applying this rule to the last example but one (art. 61) we

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arith. comp log. sin 35° = 0.241409

Sum rejecting 10 from characteristic 1.417500=log. 26.15 Applying it to the last example of Art. 61 we have

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0.000000

*ar. comp log. 101o

sum rejec. 10=2.062469 = 115.4

Applying logarithms to the example in art. 40, we have 122 + 162 = 400

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divide by 2, quot. 1.301030 = log. v400

=

From the tables 1.301030 is found to be the log. of 20 the value of the hypothenuse required.

63. We shall finish the subject of right angled triangles by presenting a case of their practical application which is likely often to occur.

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At the top of a mountain whose height was known by the barometer or otherwise to be 1000 feet above the level of the sea, a ship was observed through the tube of the instrument described at Art. 10, and the number of degrees between the tube and a plumb line from the centre of the circle was found to be 77° 30'; required the distance of the ship.

A right angled triangle is here formed, in which are given the perpendicular and angle at the vertex, and the base is required.

Referring to the rule R : tan. of one of the acute angles :: side adjacent side opposite, we have in this case

:

1010 : tan. 77° 30': : 1000: dist. required.

When arith. comp. of log. of R is used, it need not be written, since it is nothing

To find the last term of a proportion, add the logarithms of the second and third terms which are the multipliers, and the arithmetical complement of the logarithm of the first term, which is the divisor.

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'Therefore the distance of the ship is 4510.71 feet, or a little over of a mile.

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In the first of these, by a proportion previously demonstrated (Art 38,) we have

whence

R: Sin B::AB:AD

RX AD Sin B X A B

Again, in the right angle triangle a C D, we have

whence

R: Sin C:: AC: AD

RXAD Sin C XAC

The first members of the two equations above, being the same, the second members are equal, hence

sin B XA B sin C X A C

Turning this equation into a proportion, by making the first product the extremes, and the second the means, we have

sin B: sin c :: AC:AB

That is, the sines of the angles of any plane triangle, bear the same proportion to each other as the opposite sides.

From the nature of the above demonstration the selection of the vertex from which to let fall the perpendicular being entirely unrestricted, it is plain that this rule applies to all the angles and sides alike.

When therefore two of the three given parts of a triangle are opposite to each other, the part opposite to the third part may be found by the proportion which has just been established.

We shall according to our custom suppose a practical problem which introduces the case of solution in question.

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the impassable marsh, for the purpose of estimating the expense of a causeway.

Plant two staves with small flags, the one at the border of the marsh opposite the town where the causeway is to terminate, and the other at some convenient point from which the first and the town may both be seen. Measure the distance between the staves, and let it be 1000 yards; observe also the angle at the second staff by turning the tin tube of the instrument for taking angles first to the staff on the border of the marsh and then to the town; let this angle be 102°; observe also the angle at the town subtended by the line of the two staves, and let this last be 30°. A triangle will be formed in which are known an angle 30°, the side opposite 1000 yards, and another angle 102°, the side opposite to which is required. Then

sin 30°: sin 102° 1000 length of causeway required.

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