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July 29, 2014

July 29, 2014

Posted by **Chris** on Sunday, May 2, 2010 at 11:33pm.

- math -
**Reiny**, Monday, May 3, 2010 at 1:01amBefore I type out a rather lengthy solution, does your book have an answer of x = 206.8 feet ?

- math -
**Daniel**, Monday, May 3, 2010 at 1:33amHi, 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.

img121.imageshack.us/i/dsc01000vu.jpg/

- math -
**Reiny**, Monday, May 3, 2010 at 8:19amnice, 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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