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Let QO = x and given that BQ = q, such that BO = q+x is the height of the wall.

Let AO = y and given that SA = q.

BAO = α and QSO = β In ∆BAO,  BO = AB sin α eq. 1  AO = AB cos α eq. 2

In ∆QSO,  QO = SQ sin β eq. 3  SO = SQ cos β eq. 4

Subtracting eq. 2 from eq. 4, we get

SO –AO = SQ cos β – AB cos α

SA = SQ cos β AB cos α

[from the above figure, SO –AO = SA = p]

p = AB cos β AB cos α

p = AB (cos β – cos α) …eq. 5

And subtracting eq. 3 from eq. 1, we get

BO –QO = AB sin α – SQ sin β

BQ = AB sin α SQ sin β

[from the above figure, BO –QO = BQ = q]

q = AB sin α AB sin β

q = AB (sin α sin β) eq. 6

Dividing eq. 5 and eq. 6, we get  Hence, proved.

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