Posted by Chris on Sunday, May 2, 2010 at 11:33pm.
Before I type out a rather lengthy solution, does your book have an answer of x = 206.8 feet ?
Hi, this is a worksheet problem so i do not have the answer to it. It came with a pic though. Here is the link. Thanks.
nice, I had appr. the same figure,
let's call the other angle b
then tanb = 145/x
tan(Ø+b) = 295/x
we know tan(Ø+b - b) = (tan(Ø+b) - tanb)/(1 + tan(Ø+b)tanb)
tanØ = (295/x - 145/x) / ( 1 + (295/x)(145/x))
= (150/x) / (x^2 + 42775)/x2)
= 150x/(x^2 + 42775)
sec^2Ø dØ/dx = [ (x^2+42775)(150 - 2x(150x) ]/(x^2 + 42775)^2
= 0 for a max/min of Ø
[ (x^2+42775)(150 - 2x(150x) ]/(x^2 + 42775)^2 = 0
(x^2+42775)(150 - 2x(150x) = 0
150x^2 + 6416250 - 300x^2 = 0
x^2 = 42775
x = 206.8
then tanØ = (150(206.8)) / (206.3^2 + 42775)
tan Ø = .362632
Ø = 19.9° or 3479 radians
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