858
THEORY AND PRACTICE OP ENGINEERING.
Book* II.
90°, the remainder will be the unknown angle; to find theproceed by the rule of three, placing in the first term 10,000,in the second the value of the known hypotenuse, and* inthe third the sine of the angle opposite to the side required:the quotient will give the unknown side. As to find thelength A B by forming a triangle A B C, of which the hy-potenuse C B is 108 feet, the acute angle A C B 50°, andthe right angle CAB 90°. To find the unknown angleC B A, subtract the ac.ute angle A C B, 50°, from 90°, theremainder, 40°, will be the angle C B A ; then place for thefirst term 10,000, in the second the length of the hypotenusereduced to seconds, as 15552, and the third term the sine ofthe angle A C B 50°, which sought for in the tables is foundto be 76604 ; the quotient will give 11913 seconds, and the
two other unknown sides,
Fig. 1081.
Fig. 1082.
remainder 45408 is about one half second more, consequently the length of A B is 82 feet,4 inches, 8 seconds, 9i thirds. To obtain the side A C , you must adopt the same method.
Another example may he given .* where the. length of the hypotenuse is not known .— From9C° subtract the known acute angle, the remainder will be theunknown angle; to find the length of the unknown sidewhich forms the right angle and is a tangent, place in thefirst term, as before, 10000, in the second the known side, inthe third the tangent of the angle opposite to the requiredangle : the quotient will be the length. The height A C oftile triangle A B C, of which the side A B is 11913*5 seconds,the angle C B A 40°, and CAB 90°, it is required to find theheight A C ; subtract from 90° the acute angle 40°, and theremainder 50° will be the acute angle ACB. Then placein the first term 10,000, in the second 11913*5, and in thethird the tangent of the angle, 40°, which will be 83909, ac-cording to the tables ; the quotient gives 9996 seconds for theheight A C , which reduced to feet is 75 feet 5 inches. If the length of the hypotenuse isrequired, place in the first term 10000, in the second the length of the side A B, 11913*5.and in the third place the secant of the angle, 40°, which is 130540 in the table, and thequotient gives 15551, with a remainder of 88290, or nearly one-third of a second, conse-quently the hypotenuse is in length 1555l£ seconds, or 108 feet nearly.
Required the length A B of the third triangle, having formed the right-angled triangleABC, of which the hypotenuse CB is 108 feet, 15551 ^ seconds, and the side C A is69 feet 5 inches. First find the acute angle A C B opposite the required side A B by theunknown angle C B A, which you can obtain by the simple w A
rule of three, placing in the first term the value of the hypo-tenuse C B, in the second term 10000, and in the thirdterm the length of the side A C ; the quotient will give 64277for the sine, which being sought in the table and not found,seek the nearest number to it, 64278, which will give 40° forthe angle C B A ; subtract from 90°, the angle C B A, 40°, andthe remainder, 50°, will be the angle A C B. To find the sideA B, adopt the rule given for the first triangle, that is to say,by the rule of three, placing in the first term 10000, in thesecond the length of the hypotenuse, and in the third thesine of the angle ACB 50, opposite the required side A B, *‘S* 1083 *which sought in the tables will be 76604 ; the quotient of the sum will give 11913 seconds ;there will remain 45408, which is half a second, and which reduced into feet gives 82 feet,8 inches, and 9£ seconds.
Required the distance from B to C» hating formed the right-angled triangle ABC, of whichthe side A B is 82 feet, 8 inches, seconds, or 11913 Aseconds; and the side A C is 69 feet, 5 inches, and haffa second, and the angle B A C a right angle of 90°.
It will be seen that since we do not know the hypote-nuse, B C, we must work by tangents and secants.
To find the required side B C, we must know one ofthe two acute angles, as C B A, opposite to the shortside A C , 9996J seconds. To find this by proportion,place in the first term the length of the side A B, 11913*seconds; in the second term 10000. and in the third
Fig. 1084.