Question

A ladder of length 6 m makes an angle of 45° 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 the opposite wall of the room, it makes an angle of 60° with the floor. Find the distance between two walls of the room. ( in metres)​

Answer 1

Answer:

answer is 7.24m

Step-by-step explanation:

the length of the ladder is 6m.

See the figure

on leaning against the wall it first makes an agle of 45° and keeping the base fixed it makes an agle of 60° on the opposite wall.

To find the distance between the 2 walls we need to use the formula of cos theta as in figure a right angle is formed according to question.

for length QR

$\cos( {45}^{o} ) = \frac{QR}{6} \\ \frac{1}{ \sqrt{2} } = \frac{QR}{6} \\ \frac{1}{ \sqrt{2} } \times 6 = QR \\ QR = 4.24$

for the length PQ

$\cos( {60}^{o} ) = \frac{PQ}{6} \\ \frac{1}{ 2 } = \frac{PQ}{6} \\ \frac{1}{ 2 } \times 6 = PQ \\ PQ = 3$

therefore on adding PQ+QR = 4.24m + 3m

=7.24m,

Answer 2

Answer :

The distance between two walls of the room = 7.23m

Step-by-step Explanation : (please see the attached document for diagram)

Given :1) length of Ladder = 6m . makes an angle of 45° with the floor

while leaning against one wall of a room.

2) Angle made by ladder when it is fixed on the floor and made to

lean against opposite wall of the room is 60°

To find : Distance between two walls of the room = ?

Let AP and DP be the position of the ladder whose length is 6 m.

In rt. ∆ABP, BP / AP = cos60°

BP / 6 = 1/2

BP = 1/2 × 6 = 3

BP = 3m

In rt. ∆DCP , PC / DP = cos 45°

PC / 6 = 1/√2

PC = 6/√2

PC = 6/√2 × √2/√2. (Multiplying and dividing by √2 on R.H.S)

PC = 3√2 m

Distance between two walls,

= BP + PC = 3 + 3√2

= 3 + 3 × 1.41

= 3(1 + 1.41) = 7.23 m