A 10,m long ladder is placed in such a manner | vedclass

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(A) Let the length of the ladder be $L = 10\,m$ and the distance from the wall be $d = 5\,m$.Let $	heta$ be the angle with the floor and $\phi$ be th

Vedclass · Accepted Answer

$30^\circ$ with wall,$60^\circ$ with floor

Vedclass · Answer

(A) Let the length of the ladder be $L = 10\,m$ and the distance from the wall be $d = 5\,m$.Let $	heta$ be the angle with the floor and $\phi$ be the angle with the wall.In the right-angled triangle formed by the ladder,the wall,and the floor,$\cos(	heta) = \frac{	ext{base}}{	ext{hypotenuse}} = \frac{5}{10} = \frac{1}{2}$.Therefore,$	heta = \cos^{-1}(\frac{1}{2}) = 60^\circ$.Since the sum of angles in a triangle is $180^\circ$ and one angle is $90^\circ$,the angle with the wall is $\phi = 90^\circ - 60^\circ = 30^\circ$.Thus,the ladder makes an angle of $30^\circ$ with the wall and $60^\circ$ with the floor.

$A$ $10\,m$ long ladder is placed in such a manner that its lower end remains $5\,m$ away from the base of a wall. Find the angles made by the ladder with the wall and with the floor.

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