Answer :

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


[ SQ=AB=length of the ladder]


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


[ SQ=AB=length of the ladder]


q = AB (sin α sin β) eq. 6


Dividing eq. 5 and eq. 6, we get




Hence, proved.


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