A ladder rests against a wall at an angle to the horizontal. Its foot is pulled away from the wall through a distance so that it slides a distance down the wall making an angle of with the horizontal, then

A D BC E The length of the ladder is given by and the final position is given by . Length of is "" and length of is "".Let the length of the ladder be "" . Now, in ,⇒ = ⇒ = ⇒ = + + ⇒ = ⇒ = − [equation I ] Now,⇒ = + ⇒ = ⇒ = [equation II ]⇒ = Now, In , ⇒ = ⇒ = [As is also the length of the ladder which is assumed as "" ]⇒ = ⇒ + = [Since, in the diagram, = + ]⇒ + = [ Given: = ]⇒ = − [equation III ] Now, =⇒ = ⇒ = [equation IV ] Now, divide equation III by equation I, we get, − = − Substitute the values of and from equation II and IV, − ⇒ = − ( − )⇒ = ( − ) ( − )⇒ = ( − ) Now, we will be using the formulas of − and − to proceed further. + ⇒ − = 2 (2 − − ) ( )2 + ⇒ − = 2 (2) ( )2 Therefore, 2 + − ⇒ = 2 2 2 − + After cancelling, ( 2 ) ( 2 ) + (⇒ =2 ) + ( 2 ) + ⇒ = ()2 + ⇒ = ( )2

A ladder rests against a wall at an angle to the horizontal. Its foot is pulled away from the wall through a distance so that it slides a distance down the wall making an angle of with the horizontal, then

A D BC E The length of the ladder is given by and the final position is given by . Length of is "" and length of is "".Let the length of the ladder be "" . Now, in ,⇒ = ⇒ = ⇒ = + + ⇒ = ⇒ = − [equation I ] Now,⇒ = + ⇒ = ⇒ = [equation II ]⇒ = Now, In , ⇒ = ⇒ = [As is also the length of the ladder which is assumed as "" ]⇒ = ⇒ + = [Since, in the diagram, = + ]⇒ + = [ Given: = ]⇒ = − [equation III ] Now, =⇒ = ⇒ = [equation IV ] Now, divide equation III by equation I, we get, − = − Substitute the values of and from equation II and IV, − ⇒ = − ( − )⇒ = ( − ) ( − )⇒ = ( − ) Now, we will be using the formulas of − and − to proceed further. + ⇒ − = 2 (2 − − ) ( )2 + ⇒ − = 2 (2) ( )2 Therefore, 2 + − ⇒ = 2 2 2 − + After cancelling, ( 2 ) ( 2 ) + (⇒ =2 ) + ( 2 ) + ⇒ = ()2 + ⇒ = ( )2

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A ladder rests against a wall at an angle 𝛼 to the horizontal. Its foot is pulled away from the wall through a distance 𝑎 so that it slides a distance 𝑏 down the wall making an angle of 𝛽 with the horizontal, then

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𝑎 = 𝑏 𝑡𝑎𝑛

𝑎 = 𝑏 𝑐𝑜𝑡

𝑎 = 𝑏 𝑡𝑎𝑛
𝛼+𝛽

2𝛼+𝛽2𝛼−𝛽

None of the above

answer is A.

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Detailed Solution

A
   D
    B
C E
The length of the ladder is given by 𝐴𝐶 and the final position is given by 𝐷𝐸. Length of 𝐴𝐷 is "𝑏" and length of 𝐶𝐸 is "𝑎".
Let the length of the ladder be "𝑙" 𝑚. Now, in 𝛥𝐴𝐵𝐶,
𝐴𝐵
⇒ 𝑠𝑖𝑛 𝛼 = 𝐵𝐶
𝐴𝐵
⇒ 𝑠𝑖𝑛 𝛼 = 𝑙
⇒ 𝑠𝑖𝑛 𝛼 =

𝐴𝐷 + 𝐷𝐵
𝑙
𝑏 + 𝐷𝐵
⇒ 𝑠𝑖𝑛 𝛼 = 𝑙
⇒ 𝑏 = 𝑙 𝑠𝑖𝑛 𝛼 − 𝐷𝐵 [equation I ] Now,

⇒ 𝑙 𝑠𝑖𝑛 𝛼 = 𝑏 + 𝐷𝐵
   𝐵𝐶
⇒ 𝑐𝑜𝑠 𝛼 = 𝐴𝐶
𝐵𝐶
⇒ 𝐵𝐶 = 𝑙 𝑐𝑜𝑠 𝛼 [equation II ]

⇒ 𝑐𝑜𝑠 𝛼 = 𝑙
Now, In 𝛥𝐷𝐵𝐸,
  ⇒ 𝑐𝑜𝑠 𝛽 = 𝐵𝐸
𝑙

𝐵𝐸
⇒ 𝑐𝑜𝑠 𝛽 = 𝐷𝐸
[As 𝐷𝐸 is also the length of the ladder which is assumed as "𝑙" ]
⇒ 𝐵𝐸 = 𝑙 𝑐𝑜𝑠 𝛽
⇒ 𝐵𝐶 + 𝐶𝐸 = 𝑙 𝑐𝑜𝑠 𝛽 [Since, in the diagram, 𝐵𝐸 = 𝐵𝐶 + ]
⇒ 𝐵𝐶 + 𝑎 = 𝑙 𝑐𝑜𝑠 𝛽 [ Given: 𝐶𝐸 = 𝑎 ]
⇒ 𝑎 = 𝑙 𝑐𝑜𝑠 𝛽 − 𝐵𝐶 [equation III ]
Now, 𝑠𝑖𝑛 𝛽 =

𝐷𝐵

𝐷𝐸
⇒ 𝑠𝑖𝑛 𝛽 = 𝐷𝐵
𝑙
⇒ 𝐷𝐵 = 𝑙 𝑠𝑖𝑛 𝛽 [equation IV ]
Now, divide equation III by equation I, we get,
𝑎 𝑙 𝑐𝑜𝑠 𝛽 − 𝐵𝐶
𝑏 = 𝑙 𝑠𝑖𝑛 𝛼 − 𝐷𝐵
Substitute the values of 𝐵𝐶 and 𝐷𝐵 from equation II and IV,
𝑎 𝑙 𝑐𝑜𝑠 𝛽 − 𝑙 𝑐𝑜𝑠 𝛼
⇒ 𝑏 = 𝑙 𝑠𝑖𝑛 𝛼 − 𝑙 𝑠𝑖𝑛 𝛽
𝑎 𝑙 (𝑐𝑜𝑠 𝛽 − 𝑐𝑜𝑠 𝛼)
⇒ 𝑏 = 𝑙 (𝑠𝑖𝑛 𝛼 − 𝑠𝑖𝑛 𝛽)
𝑎 (𝑐𝑜𝑠 𝛽 − 𝑐𝑜𝑠 𝛼)
⇒ 𝑏 = (𝑠𝑖𝑛 𝛼 − 𝑠𝑖𝑛 𝛽)
Now, we will be using the formulas of 𝑐𝑜𝑠 𝐴 − 𝑐𝑜𝑠 𝐵 and 𝑠𝑖𝑛 𝐴 − 𝑠𝑖𝑛 𝐵 to proceed further.
𝐴 + 𝐵
⇒ 𝑐𝑜𝑠 𝐴 − 𝑐𝑜𝑠 𝐵 = 2 𝑠𝑖𝑛 (
2
𝐴 − 𝐵

𝐵 − 𝐴
) 𝑠𝑖𝑛 ( )
2
𝐴 + 𝐵
⇒ 𝑠𝑖𝑛 𝐴 − 𝑠𝑖𝑛 𝐵 = 2 𝑠𝑖𝑛 (
2

) 𝑐𝑜𝑠 ( )
2
Therefore,

𝑎 2 𝛽 + 𝛼 𝛼 − 𝛽
⇒ = 2 2 𝑏 2

𝛼 − 𝛽

𝛼 + 𝛽
After cancelling,

𝑠𝑖𝑛 ( 2 ) 𝑐𝑜𝑠 ( 2 )
𝛽 + 𝛼
𝑎 𝑠𝑖𝑛 (
⇒ 𝑏 =

2 )
𝛼 + 𝛽
𝑐𝑜𝑠 ( 2 )
𝑎 𝛼 + 𝛽
⇒ 𝑏 = 𝑡𝑎𝑛 (

)
2
𝛼 + 𝛽
⇒ 𝑎 = 𝑏𝑡𝑎𝑛 ( )
2

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