860
THEORY AND PRACTICE OK ENGINEERING.
Book II.
adjacent side are known. This method is almost the same as that for the fifth triangle,only instead of taking the sine of the ohtuse angle, we take the sine of its complement, aswill be seen in the following example. Required thedistance from A to B, having formed the obtuse-angledtriangle ABC, of which the angle CAB 83°, ACB109°, and the adjacent side AC 52 feet. To find theunknown angle C B A, add together the value of thetwo known angles CAB 33°, and ACB 109°, andsubtract their sum, 142°, from 180°; the remainder,
38°, will be the value of the unknown angle CBA.
To find the length of the unknown side A B by therule of three, place in the first term the sine of the angle C B, 38°, opposite the known sineA C ; this sine will be 61566 ; in the second term the value of the known side AC 52,and in the third term the sine of the angle A C B 109, opposite the side required. Butsince this angle A C B is 109° and we have no sine above 90° we must take the sine of thecomplement of 109°; that is to say, we must subtract the obtuse angle ACB 109° from180°; the remainder will be 71°, and the sine of 71° will be 94551, which must be put inthe third term ; the quotient will give 79 with a remainder for the length A B.
It will be seen that all the other difficulties which might attend the obtuse-angledtriangle are reduced to the rules for the acute-angled triangle, provided it is remem-bered that to have the sines of obtuse-angled triangles the complement of 180° mustbe taken.
To find the three Angles of a Triangle , when the three sides are known, by the calculationof sines.— Since this difficulty may occur in the three kinds of triangles, right, acute, andobtuse-angled we will commence witli the first, viz. the right-angled. In the triangle ABC, the three angles of whichwe wish to ascertain, let it be supposed that the side A B is11913 seconds, AC 9996, and CB 15552; to find the threeangles, let fall a perpendicular A I) from the angle C A B tothe side CB; add together the two sides AB 11913 andAC 9996; their sum will be 21909; subtract the shorterside A C from the mean A B. to have their difference 1917 :then by proportion, placing in the first term the greatest sideC B, 15552, in the second the sum of the two sides A B andAC , 21909, and in the third term the difference 1917; thequotient will give 2700, with a remainder of 9153, which is j-ig. |089.nearly two-thirds of the difference between the two segments
C D and D B, and this difference being subtracted from the half length C B leaves12851^, of which take half for the small segments CD, and if to this half, 6425.^, you addthe difference, 2700J, between the two segments, you will have for the other segment, D B,9125.^. Before proceeding further, value the small portions of the two segments C D andD B, and it will give 6425^ for CD, and 9126^ for DB. To find the angles of the tworight-angled triangles, ADC and A B D, of which we know the hypotenuse and oneside, follow the rule of the third triangle. To find the two acute angles 1) A C and D C Aof the right-angled triangle ADC; by proportion, place in the first term the hypotenuseAC , 9996, in the second 100000, and in the third the side CD, 6425^; to have theopposite angle D A C , the quotient will give 64282, which will be a sine; seek the nearestnumber in the table, 6427b, which will give 40° for the angle D A C, which subtractedfrom 90° leaves 50° for the other angle DCA: by the same rule it will be foundthat the angle D A B in the right-angled triangle ABD is 50° and DBA 40°: addthese together; their sum will be 90° for the third angle CAB of the right-angledtriangle.
Mensuration of Distances, Heights , §*c. by Logarithms , which facilitates the calculations ofsines, and avoids the necessity of calling in the aid of proportion, addition, and subtractionwith them, hold the place of multiplication and division, as has been seen in precedingexamples ; here we shall repeat them, to show the convenience of logarithms in calculatingsines, tangents, and secants. We have already mentioned that sines give halves, thirds,quarters, &c., by reason of their remainder, after their division; logarithms only giveintegers, which obliges us to reduce the value of the known sides into small measures, as faras 4000 (the number given in the ordinary tables), which is a sufficient quantity to beappreciated in mensuration.
Measuring Lengths , by forming a right-angled triangle, of which the hypotenuse, withthe right and acute angles are known. To find the third angle, subtract from 90° thevalue of the known acute angle; the remainder is the third unknown angle: then to findthe side opposite to the acute angle just ascertained, seek in the logarithms in the columnof numbers, the number of the length of the hypotenuse figured apart. Then seek inthe table of sines, in the column of logarithmic sines, the logarithmic sine of the acute