line CF, (till you see the staff E and object B in the same direction,) say 45 yards. Now it is manifest, that a perpendicular from E falling on the line CF, will be equal to AC, viz. 62 yards; and that CD, the distance between said perpendicular and AC, will be equal to AE, viz. 20 yards: And if we subtract CD, or 20 yards, from CF, or 45 yards, the remainder will be FD, or 25 yards. Now, because the triangle BCF is similar to EDF, the similar sides are directly proportional. Therefore, FD is to DE as FC is to CB; or 25 is to 62 as 45 is to CB. By this statement CB is found to be equal to 111.6 yards. Therefore, AB, the required breadth of the impassable river, is 111.6-62, or 49.6 yards. Let AE be 39.1 feet, CF 64.1, and DE 39 feet; AB is required. Ans. 100 feet, nearly. To find the height of a tree or steeple, if the sun be above the horizon, and its shadow can be measured; set up a pole perpendicular to the horizon, and measure the length of its shadow. Then say :-As the length of the shadow of the pole is to its perpendicular height, so is the length of the shadow of the tree, or of any object, to its perpendicular height. EXAMPLE. If a pole 10 feet long cast a shadow 8 feet in length, what is the height of an object the shadow of which is 50 feet long? Ans. 621 feet. Another method is by means of two poles of unequal lengths, set up parallel to the object, so that the observer may see the top of the object over the tops of both poles. Thus, let the length of the pole dr be 6 feet, that of the pole sg 8 feet, their distance rg 10 feet, and the distance rC of the shorter pole from the object AC, 190 feet. Then, because the triangles sdn and ɗKA are similar, dn is to ns as dK to KA; or 10: 8-6:: 190 to AK. Hence, AK=38 feet; and AC=38+6=44 feet. If the eye of the observer be placed very near the earth, then only one pole will be required. For if the eye of the observer be placed at B, and the top of the object AC be just visible over the top of the pole dr, then Br: rd :: BC : AC. Let BC equal 100 feet, Br 8 feet, and rd 6 feet; required the height of the object AC. Ans. 75 feet. Let Bg equal 12 feet, gs 8 feet, and BC 90 feet; required the height of AC. Ans. 60 feet. In all triangles, the sides are to each other as the sines of the angles opposite the said sides; and any given side is to the sine of the angle opposite said side, as any other given side is to the sine of its opposite angle. Thus, (in Fig. 2,) the sine of the angle at A is to the base BC, as the sine of the angle at B to the side AC. Or, the sine of the angle at C is to the side AB, as the sine of the angle at A is to the base BC. And the side BA is to the sine of the angle at C, as the side AC is to the sine of the angle at B. In any right-angled triangle, if either of the acute angles be given, the other may be found by subtracting the given angle from 90 degrees; and in all cases, the sum of any two angles subtracted from 180 degrees, will give the third angle. Thus, if the angle at B (Fig. 2) contain 35 degrees, the angle at A will equal 90-35, or 55 degrees; and if the angle at a [see the triangle ¶ 19] is 46 degrees, and the angle at b is 70 degrees, then the angle at c equals 180-(46+70,) or 64 degrees. The natural sines corresponding to any number of degrees, and likewise the number of degrees corresponding to any natural sine, may be found in the table of natural sines, ¶ 73. EXAMPLES. 1. Given, the base of the right-angled triangle ABC, viz. BC 96 rods, and the angle at B (between the base and the hypotenuse) 18 degrees; required the perpendicular AC. Ans. 31.192 rods. The angle at A equals 90-18=72 degrees; and in the table (73) we find the sines of 72 and 18 degrees to be .951057 and .309017; therefore, as .951057: 96:: .309017 to AC. 2. If the perpendicular of the right-angled triangle EBA (Fig. 1) be 40 rods, and the angle at E is 70 degrees, what I will be the breadth of the river AB? Ans. 109.89 rods. 3. If the base of a triangle is 60 feet, and the angles at the extremities of the base are 50 and 60 degrees; required the other two sides. Ans. 55.296 and 48.912 feet. 4. The base of a right-angled triangle is 222 feet, and the hypotenuse 423 feet; required the two acute angles. = Ans. 31° 40′ and 58° 20'. Thus, as 423: 1 :: 222 to the sine of the angle opposite the base, viz. 0.524822; and .524822 minus .515038, the sine of 31 degrees, .009784; and .529919, the sine of 32 degrees, minus .515038=.014880; then, as .014880 : 60 :: 0.009784 to the number of minutes which the angle opposite the base exceeds 31 degrees; and consequently, the angle equals 31 degrees, 40 minutes, nearly; and 90-31° 40′: 58° 20'. = The complement of an angle is what it wants of 90 degrees, and the supplement of any angle is what it wants of 180 degrees. Thus, the supplement of 106 degrees is (180-106), or 74 degrees; and the sine of any obtuse angle is the same as the sine of its supplement. 5. The angle opposite to the diagonal of a trapezium is 106 degrees, (see 20,) and the angle cba (viz. the angle at b) is 38 degrees, and the side opposite this angle, viz. ac, is 53 rods; the diagonal ab is required. Ans. 82.75 rods. As .615661 53: 0.961262 (the sine of 106, or of 74 degrees) to the side required. See ¶ 81, prob. 132. 64. IMPORTANT GEOMETRICAL RATIOS. The circumference of a circle, or of a sphere, whose diameter is 1, is 3.141592654 ... The area of a circle whose diameter is 1, is .78539816, (or one-fourth the circumference of the circle;) and the area or surface of a sphere is four times that of a circle of the same diameter. The areas of squares are as the squares of their sides; and the areas of circles are as the squares of their diameters; and the areas of similar triangles are as the squares of their corresponding or homologous sides; and the areas of all regular polygons are as the squares of their sides. The area of the circle is to that of its circumscribing square as .785398 to 1; and the area of an ellipse bears the same ratio to its circumscribing rectangle. The triangle is one-half and the parabola is two-thirds of the circumscribing rectangle. The SOLIDITIES of all similar solids are to each other as the cubes of their sides, or diameters. Similar cones or similar pyramids are to each other as the cubes of their bases, and likewise as the cubes of their altitudes. A cylinder bears the same ratio to its circumscribing square prism, that the circle does to its circumscribing square, viz. that of .7854 to 1. A cone is one-third of its least circumscribing cylinder; and a pyramid is one-third of its least circumscribing prism of the same number of sides. A sphere is .52359878 of its circumscribing cube, and twothirds of its circumscribing cylinder. A spheroid is two-thirds of the circumscribing cylinder, whose length is equal to that of the polar axis of the spheroid. The paraboloid is one-half of its circumscribing cylinder. The parabolic spindle is eight-fifteenths of its circumscribing cylinder. ¶ 65. RULES FOR THE SOLUTION OF A VARIETY OF DIF. FICULT GEOMETRICAL PROBLEMS. 1. The base of a right-angled triangle and the sum of its three sides being given, to find the perpendicular: From the sum of the three sides subtract the base, and from the square of the remainder subtract the square of the base, and divide the LAST by TWICE the FIRST remainder. EXAMPLE. The height of a tree standing upon a plane is 80 feet; at what distance from the ground must it be broken off, that the top may strike the plane 20 feet from the stump, whilst the other end rests on the top of the stump? Ans. 26 feet. 2. Given the sum of two numbers and the difference of their squares, to find those numbers :— Divide the difference of their squares by their sum, and the quotient will be their difference; add half of this difference to half the sum for the greater, and subtract half of this difference from half the sum for the less number. EXAMPLE. The sum of the sides of a rectangle is 64, and the difference between the squares of the length and breadth is 256 rods; required the length of the sides. Ans. 20 and 12 rods. 3. Given the difference of two numbers and the difference of their squares, to find the numbers : -- Divide the difference of their squares by the difference of the numbers, and the quotient will be their sum; and half their difference, added to and subtracted from half their sum, will give the numbers. |