# Homework Help: Sound attenuation barrier height (trigonometry)

1. Nov 9, 2012

### jhmz

1. The problem statement, all variables and given/known data

http://img543.imageshack.us/img543/4598/photosfr.jpg [Broken]

Find dss, dsr and thus hs.

2. Relevant equations

trig

3. 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)

Last edited by a moderator: May 6, 2017
2. Nov 9, 2012

### DrOnline

http://img543.imageshack.us/img543/4598/photosfr.jpg [Broken]

OP's image.

Last edited by a moderator: May 6, 2017
3. Nov 9, 2012

### haruspex

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.