Sound attenuation barrier height (trigonometry)

AI Thread Summary
The discussion focuses on solving a trigonometric problem related to sound attenuation barrier height, specifically finding distances dss and dsr using given equations. The user attempts to express these distances in terms of the height hs and angles θ and ∅, utilizing trigonometric identities. The challenge arises from the complexity of substituting equations and solving for hs, particularly with the constraints provided by ISO9613 regarding path differences. The user also outlines a method to derive a quadratic equation involving height, indicating a systematic approach to the problem. Overall, the thread highlights the intricacies of applying trigonometry to real-world noise assessment scenarios.
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Homework Statement



http://img543.imageshack.us/img543/4598/photosfr.jpg

Find dss, dsr and thus hs.

Homework Equations



trig

The Attempt at a Solution


dss = hs/(sin[θ])
dsr = hs/(sin[∅])

dss = 3.059/(cos[θ])
dsr = 47.191/(cos[∅])

θ = arctan(hs/3.059)
∅ = arctan(hs/47.191)

I have tried many times to solve this; usually by substituting the last two equations into the first 4 and trying to solve but without success.

The problem is to do with ISO9613 where Z (path difference between dss+dsr and d) = 7.81 m and d = 50.25 m.

http://www.cevreselgurultu.cevreorm...ps/assessment_methods/industry_ISO_9613-2.pdf (page 21-25)
 
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http://img543.imageshack.us/img543/4598/photosfr.jpg

OP's image.
 
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Let d comprise d1 (dss side) and d2. Write D = dss+dsr.
D = √(d12+h2) + √(d22+h2)
D2 = d12+2h2+d22+2√((d12+h2)(d22+h2))
D2 - d12-2h2-d22=2√((d12+h2)(d22+h2))
Squaring again leaves a quadratic in h2.
 

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