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SOLUTION OF RIGHT ANGLED TRIANGLES, WITH THE AID
EXAMPLE. 61. Referring to the example of Art. 39, where the hypothe
1010 X 15
sin 350 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
log. of 1010 = 10.000000 add log
15 = 1.176091
Remainder 1.417500 - = log. of 26.15 = the hyp.
Referring to Art. 41 where the height of the tower c is
C = tan 300 x 200 which becomes, when the radius 1 in the first term of the proportion is changed into 1010
tan 300 x 200
1010 we have, by applying logarithms
log. tan 30o = 9.761439 add log. 200 = 2.301030
12.062469 Subtract log. 101° = 10
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.
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
log a + 10 log 6 subtract 10 and there remains
log a The same result as would have been obtained by subtracting log. b from log. a.
Hence to perform operations containing a number of multiplications and divisions, by means of logarithms, we have the following
Write the logarithms of all the multipliers, and the arith. coinplements 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. log. 15
Applying this rule to the last example but one (art. 61) we have
log. 10° = 10.000000
1.176091 arith. comp log. sin 35o = 0.241409 Sum rejecting 10 from characteristic 1.417500= log. 26.15 Applying it to the last example of Art. 61 we have
log. tan. 300 9.761439
log. 200 2.301030 *ar. comp log. 1019 0.000000
sum rejec. 10 = 2.062469 Applying logarithms to the example in art. 40, we have 12? + 16 = 400
log. of 400 = 2.602060 divide by 2, quot. = 1.301030 = log. 1400 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.
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
1000 = 3.000000
log. 4510.71 feet. Therefore the distance of the ship is 4510.71 feet, or a little over of a mile.
SOLUTION OF PLANE TRIANGLES IN GENERAL.
64. Let ABC be any tri
A angle. From the vertex of one of the angles a, let fall upon the side opposite, the perpendicular Ad; the given triangle will be divided into two right angled triangles B D ABD and ACD.
In the first of these, by a proportion previously demonstrated (Art 38,) we have
R: sin B:
:: AB: A D whence
RXAD= sin B XA B
R: sin c::AC:AD whence
RX AD=sin cXA C The first members of the two equations above, being the same, the second members are equal, hence
sin B XAB= sin c XAC 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: A B
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 he above demonstration 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.
Let it be required to ascertain accurately the
distance, from the town in the figure, across 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 30o. 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 300 : sin 102° : : 1000 : length of causeway required.