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A ladder sets against a wall at an angle α to the horizontal. If the foot is pulled away from the wall through a distance of 'a', so that is slides a distance 'b' down the wall making an angle β with the horizontal. Show that cosα - cos β/sin β - sin α = a/b
Ans: Let CB = x m. Length of ladder remains same
Cos α = CB/CA ∴ ED = AC Let Ed be
Cos α = x/h ∴ ED = AC = h
x = hcos α .........(1)
cos β =DC + CB/ED
cos β = a + x/h
a + x = hcos β
x = hcos β - a .........(2)
from (1) & (2)
hcos α = hcos β - a
hcos α - hcos β = - a
-a = h(cosα - cosβ) .........(3)
Sin α = Sin α = AE + EB/AC
Sin α =b + EB/ h
hSin α - b = EB
EB = hSin α - b .........(4)
Sin β = EB /DE
Sin β = EB/h
EB = hSin β .........(5)
From (4) & (5)
hSin β = hSin α - b
b = hsin α - hSin β
-b = h(Sin β - Sin α) .........(6)
Divide equation (3) with equation (6)
- a/-b = h(cosα - cos β )/ h(sin β - Sinα )
∴ a/b = Cosα - Cosβ/Sinβ - Sinα
integral 0 to 4 integral 0 to y root of 9+ysquredxdy
proof of chebychevs lemma
4 1/2 ----2----1/3=3
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