Question

A ladder 6m long leand against a wall at the height 4.8 find the distance between the wall and the foot of the ladder.​

Answer 1

Step-by-step explanation:

by pithagoreas theorm

BC^2=AB^2+AC^2

BC^2=4.8^2+6^2

BC^2=23.04+36

BC^2=59.04

BC=√59.04

BC=7.683

Answer 2

7.6837 is the answer.

The ladder is of 6 m.

The building is 4.8 m high.

We have to find the distance between the wall of the building and foot of ladder.

This can form a triangle ABC.

By Pythagoras theorem,

${ab}^{2} + {bc}^{2} = {ac}^{2}$

${6}^{2} + {4.8}^{2} = {ac}^{2}$

$36 + 23.04 = {ac}^{2}$

${ac}^{2} = 59.04$

$ac = 7.6837$

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