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p>A Ladder makes an angle of $60^{o}$ with the ground when placed against a wall. If the foot of the ladder is 2 m away from the wall, then the length $( in\ meters)$ is:
( A) \ \frac{4}{\sqrt{3}}$
$( B) \ 4\sqrt{3}$
$( C) \ 2\sqrt{2}$
$( D) 4$
Given: Angle made with the ground$\displaystyle =60^{o}$. Distance of the foot of the ladder from the the wall$\displaystyle =2\ m$
To do: To find the length of the ladder.
Solution:
In the figure, MN is the length of the ladder, which is placed against the wall AB and makes an angle of $60^{o}$ with the ground.
The foot of the ladder is at N, which is 2 m away from the wall.
$\therefore BN\ =\ 2\ m$
In right-angled triangle $\vartriangle MNB$:
$cos60^{o} =\frac{BN}{MN} \ =\frac{2\ m}{MN}$
$\Rightarrow \frac{1}{2} =\frac{2\ m}{MN}$
$\Rightarrow MN=4\ m$
Therefore, the length of the ladder is $4\ m$.
Hence, The correct optn is $( D)$.
Updated on: 10-Oct-2022
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