A complete and practical solution book for the common school teacher . FIG. 28. 166 FAIRCHILDS SOLUTION BOOK. (1) (2) (3) (4) Solution. Let BD^, the height of the col-umn; CD=<^, the height of thestatue; and .#=AB, the distancethe boy stands from the foot ofthe column. Describe a circle which shall passthrough D ano C, and tangentto AB, A being the point oftangent/. The angle DAC being an inscribedangle, is greater than if A weremoved either way on AB, for itwould be exterior to the circle. Since angle A is measured by halfangle C by half of the same arc,triangles ABD, FIG. 29. of the
A complete and practical solution book for the common school teacher . FIG. 28. 166 FAIRCHILDS SOLUTION BOOK. (1) (2) (3) (4) Solution. Let BD^, the height of the col-umn; CD=<^, the height of thestatue; and .#=AB, the distancethe boy stands from the foot ofthe column. Describe a circle which shall passthrough D ano C, and tangentto AB, A being the point oftangent/. The angle DAC being an inscribedangle, is greater than if A weremoved either way on AB, for itwould be exterior to the circle. Since angle A is measured by halfangle C by half of the same arc,triangles ABD, FIG. 29. of the arc AD, andwe have the similar (5) c : x :: x : c-\-d. Whence, x=.yj[c(c-{-d) ]. (6) Now, A is the point at which the boy can see the statue at the best advantage. Let r=120 ft, and d=&) ft.;then #=+ ft. PROBLEM 333. Find the three sides of a right-angled triangle whose perimeter is120 ft., and whose base is § of its perpendicular. Solution. (1) The base and perpendicular being as 3 to 4, the hypoth- enuse is V (32-{-42) = 5, proportionally. (2) This triangle whose perimeter is 3-}-4-r-5 = 12, is similar to the triangle whose sides are required; hence, 120-7-12,or 10 times these, are 30, 40 and 50 ft., the requiredsides. PROBLEM 334. Required the sides of a right triangle which shall contain thegreatest area under the shortest perimeter, when the square of the areashall be equal to the product of the three sides. Solution. (1) Let ;t—the base, y = the perpendicular; then, V (jr2-f-jj>2) = the hypothenuse. (2) By the conditions we must have %x2y2=zxyyj (
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