A complete and practical solution book for the common school teacher . tion. (1) Let AC = 50 =^; DB = 30 = b; BC = c, FE =x\ EB = y; CE = c—y. (2) Then by similar triangles, a : c :: x : y\ therefore,ay = ex . . (1). (3) Also, /; : : c—x\ there- fore, be—by—ex . (2). (4) Now from (1) and (2), ay-\-by — be, and y—bc-^r v . FIG. 21. (5) By substituting the value of y in the first proportion, we have x : be-r-(a+b). (6) ex — abc-1r{aJrb), and x=ab-rr{a-\-b), or . (7) Hence, (50X30)^(50+30) =18$ ft. .\ 18f ft. is the required height, or FE. Exile.—Divide the pro duet of their heights by
A complete and practical solution book for the common school teacher . tion. (1) Let AC = 50 =^; DB = 30 = b; BC = c, FE =x\ EB = y; CE = c—y. (2) Then by similar triangles, a : c :: x : y\ therefore,ay = ex . . (1). (3) Also, /; : : c—x\ there- fore, be—by—ex . (2). (4) Now from (1) and (2), ay-\-by — be, and y—bc-^r v . FIG. 21. (5) By substituting the value of y in the first proportion, we have x : be-r-(a+b). (6) ex — abc-1r{aJrb), and x=ab-rr{a-\-b), or . (7) Hence, (50X30)^(50+30) =18$ ft. .\ 18f ft. is the required height, or FE. Exile.—Divide the pro duet of their heights by their sum. NoTK.—It would be well to bear in mind that the height at whichthe lines cross is not affected by the distance the poles are apart. PROBLEM 323. Two trees are a certain distance apart; from the top of a to the footof t> is 60 ft.; from the top of b to the foot of a is 40 ft.; from the pointwhere these two lines intersect to the plane is 15 ft.: what is the heightof each tree and the distance apart? {American Mathematical Monthly.). 150 FAIRCHILDS SOLUTION BOOK. Solution. (1) From Fig. 21, let DB=*—y\ AC = x+y; FE = r = 15; CD = £ = 40; AB=A = 60; BF=^; and CE = w. (2) By similar triangles, w : c :: ze/-j-£ : x—y, or ^(^—y) = c{zv+z) ... (1). (3) z \ c :: w-|-,s : x-\-y, or {w-\-£)c=.z(x-\-y) ... (2). (4) {x-y)=z{x+y) . . (3). (5) Put ;t:-f-j/=5 and x—y=.t. (6) Then wt=zs, or z#=—— . . (4). (7) Substituting this value of w in the proportion of (2), 2s zt-\-zs „. .. , z — \ c .: -—-— : t. Dividing antecedents by — we it i have, s : c :: t-\-s : A or ^=<r(/-j-j). /nx „T, 5/ (x-\-y)(x—y) x2—y2 ._. (8) Whence, f=—= v T/V; ^ = Q J . . (o). v 5-K x-\-y-\-x—y 2x v (9) From (5), #2— 2o:=j2, whence .*=<:+VO^^8)- (10) From figure, AB2—AC2=CB2, and CD2—DB2=CB2. (11) .-. AB2—AC2=CD2—DB2,or«2—{x+y)2~b2—(x—y)\a2—b2 (12) Whence, x—- 4y a2—b2 (13) Equating the two values of x, —j = ^-|V (724~^2)
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