Question

A ladder of length 8m makes an angle 30 degree with the floor while leaning against one Wall of a room if the foot of the ladder is kept fixed on the floor and it is made to lean against opposite wall of the room it makes an angle of 45° with the floor find the distance between these two wall of the room

Answer 1

AnswEr:

Distance between these two walls is 4(√2 + √3) m.

ExplanaTion:

Given :

• AC = CE = 8 m

We know that,

Cos Ø = B/H

In ∆ABC

Cos 45 = BC/AC

=> 1/√2 = BC/8

=> BC = 8/√2

=> BC = 4√2 m

Now, in ∆CDE

Cos 30 = CD/CE

=> √3/2 = CD/8

=> CD = 4√3

Now, BD = BC + CD

=> BD = 4√2 + 4√3

=> BC = 4(√2+√3)

Therefore, distance between these two walls is 4(√2+√3) m.

Answer 2

Given :

• A ladder of length 8 m.
• Angle of 30° is formed between the ladder and the floor of the room while the ladder is leaning against the wall.
• The foot of the ladder is kept fixed [There is no change in ladder's base position]
• The ladder is then made to lean against the opposite wall making an angle of 45° with the floor.

To Find :

• Distance between the two walls of the room, BD.

Solution :

We have the length of ladder given as 8 m.

In ΔABC,

AC = 8 m = ladder length

BC = x m

$\bold{\theta\:=\:45\:^\circ}$

Using the trigonometric function cos.

We know,

$\sf{cos\:\theta\:=\:\dfrac{Adjacent\:side\:to\:\angle\:\theta}{Hypotenuse}}$

Block in the data,

$\longrightarrow$ $\sf{cos\:45\:^\circ\:=\:\dfrac{x}{8}}$

$\longrightarrow$ $\sf{\dfrac{1}{\sqrt{2}}\:=\:\dfrac{x}{8}}$

$\longrightarrow$ $\sf{\sqrt{2}x\:=\:8}$

$\longrightarrow$ $\sf{x\:=\:\dfrac{8}{\sqrt{2}}}$

$\longrightarrow$ $\sf{\dfrac{8}{\sqrt{2}}}$ $\sf{\times\:\dfrac{\sqrt{2}}{\sqrt{2}}}$

$\longrightarrow$ $\sf{x=\dfrac{8\:\sqrt{2}}{2}}$

$\longrightarrow$ $\sf{x=\:BC\:=\:4\:\sqrt{2}\:\:\:(1)}$

Now, consider ΔCDE.

CE = 8 m = ladder length

CD = y m.

$\bold{\theta\:=\:30\:^\circ}$

Use the same trigonometric function, cos.

$\longrightarrow$ $\sf{cos\:30\:^\circ\:=\:\dfrac{y}{8}}$

$\longrightarrow$ $\sf{\dfrac{\sqrt{3}}{2}\:=\:\dfrac{y}{8}}$

$\longrightarrow$ $\sf{8\:\times\:\sqrt{3}\:=\:2y}$

$\longrightarrow$ $\sf{\dfrac{8\:\sqrt{3}}{2}\:=\:y}$

$\longrightarrow$ $\sf{4\:\sqrt{3}=y\:=\:CD\:\:(2)}$

Now, we can see : BD = BC + CD

BC = x = 4√2 m

CD = y = 4√3 m

$\longrightarrow$ $\sf{BD\:=\:x+y}$

$\longrightarrow$ $\sf{BD\:=\:4\:\sqrt{2}\:+\:4\:\sqrt{3}}$

$\sf{BD\:=\:(4\:\times\:1.41)\:+\:(4\:\times\:1.73)}$

$\bold{\big[\because\:\sqrt{2}\:=\:1.41\:;\:\:\sqrt{3}\:=\:1.73\big]}$

$\longrightarrow$ $\sf{BD\:=\:(5.64)+(6.92)}$

$\longrightarrow$ $\sf{BD\:=\:5.64+6.92}$

$\longrightarrow$ $\sf{BD\:=\:12.56}$

$\large{\boxed{\bold{\purple{Distance\:between\:two\:walls\:=\:BD\:=\:12.56\:m}}}}$