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

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Solution

$B C = l a d d e r = 10 c m$
Given: $A B = 5 m$
To find $a n g l e C a n d$ angle B$
In $Δ A B C c o s t h e t a \frac{A B}{B C} = \frac{5}{10} = \frac{1}{2}$
$\Rightarrow= c o s^{- 1} [1 / 2] = 60^{o}$
$\Rightarrow\propto= 90 - θ = 30^{o}$
$∴$ ladder makes $30^{o}$ with wall and $60^{o}$ with the floor.

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

BC=ladder=10cmGiven: AB=5mTo find angleCandangle B$In ΔABCcosthetaABBC=510=12⇒=cos−1[1/2]=60o⇒∝=90−θ=30o∴ ladder makes 30o with wall and 60o with the floor.

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