. The elements of railroad engineering . ince the sum of all the angles equals P two right angles, it is evident that thesum of the two acute angles must equala right angle. Therefore, if in anyright-angled triangle one acute angle isknown, the other can be found by sub-C tracting the known angle from a rightangle. Thus, in Fig. 37, ^ ^ C is aright-angled triangle, right-angled at C. Then, the anglesA -^ B =z one right angle. If A = -f- of a right angle, B — 1— -3. = A of a right angle. 714. In any right-angled triangle, the square describedon the hypotenuse is equal tothe sum of the squares d
. The elements of railroad engineering . ince the sum of all the angles equals P two right angles, it is evident that thesum of the two acute angles must equala right angle. Therefore, if in anyright-angled triangle one acute angle isknown, the other can be found by sub-C tracting the known angle from a rightangle. Thus, in Fig. 37, ^ ^ C is aright-angled triangle, right-angled at C. Then, the anglesA -^ B =z one right angle. If A = -f- of a right angle, B — 1— -3. = A of a right angle. 714. In any right-angled triangle, the square describedon the hypotenuse is equal tothe sum of the squares de-scribed upon the other twosides. If A B r. Fig. 38, is aright-angled triangle, right. ^^angled at B, then the squaredescribed upon the hypotenuseA C is equal to the sum ofthe squares described upon thesides A B and B- C; conse-quently, if the lengths of thesides A B and B C are known,we can find the length of thehypotenuse by adding the squares of the lengths of the sidesA B and B C, and then extracting the square root of the 1 2 3 4 5 6 7 8 9 10 11 12 13 14 IS 16 17 18 19 20 21 22 23 24 23 Fig. 38 244 GEOMETRY AND TRIGONOMETRY. 150 Example. -If , //? = 3 inches, and PC — ^ inches, what is th«length of the hypotenuse A C? Solution.— 3* = 9; 4* = 1G; adding. 9 + 16 = 25. |/25 = 5. Therefore, A C= 5 inches. Ans. If the hypotenuse and one side are given, the other sidecan be found by subtracting the scjuare of the given sidefrom the square of the hypotenuse, and then extracting thesquare root of the remainder. —The side given is 3 inches, the hypotenuse is 5 inches;what is the length of the other side ? Solution.^ 3» = 9; 5« = 25. 25 — 9 = IG, and \^l6 = 4 inches. Ans. ExAMPi-K.—If, from a church steeple which is 150 feet high, a rope isto be attached to the top, and to a stake in the ground85 feet from its foot (the ground being supposed to belevel), what must be the length of the rope ? Solution.—In Fig. 39, .-/ 7? represents the steeple150 f
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Keywords: ., bookcentury1800, bookdecade1890, booksubjectrailroadengineering
