A ladder leaning against a wall inclined at an angle a to the horizontal. The top of t ladder touches the parapet. On moving its feet a meter away from the wall, the Lad makes an angle B to the horizontal and its top now touches a window. Prove that distance of the parapet from the window is a cot (alpha + beta)/2
Answers
distance of the parapet from the window is Cot((α + β)/2)
Step-by-step explanation:
A ladder leaning against a wall inclined at an angle α to the horizontal
Let say Ladder length = L
Sinα = Height of Parapet from ground / Ladder length
=> Height of Parapet from ground = L Sinα
Tan α = Height of Parapet from ground/ horizontal distance from bottom
=> horizontal distance from bottom = L Sinα/Tanα
=> horizontal distance from bottom = L Cosα
Sinβ = Height of Window / Ladder length
=> Height of Window = L Sinβ
Tan β = Height of window from ground/ horizontal distance from bottom
=> L Cosα + 1 = L Sinβ/Tanβ
=> L Cosα + 1 = L Cosβ
=> L ( Cosβ - Cosα) = 1
=> L = 1/( Cosβ - Cosα)
Height of Parapet from ground - Height of window from ground
= L Sinα - L Sinβ
= L (Sinα - Sinβ )
= (Sinα - Sinβ ) / ( Cosβ - Cosα)
= ( 2 Cos((α + β)/2)Sin((α - β)/2)) / ( -2Sin((β + α)/2)Sin((β - α)/2)
= Cot((α + β)/2)Sin((α - β)/2)) /(Sin((α - β)/2))
= Cot((α + β)/2)
distance of the parapet from the window is Cot((α + β)/2)
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