A ladder 17 m long reaches a window of a building 15 m above the ground.

Solution:

Let us draw the diagram from the given information we get a right angled triangle ABC as shown below,

Let the window be at the point A. We know that angle formed between the building and ground is always 90°.

Given: AB = 15 m and CA = 17 m

Now we will use Pythagoras theorem to find $l(B C)$.

$\therefore A C^{2}=A B^{2}+B C^{2}$

Let us substitute the values we get,

$\therefore 17^{2}=15^{2}+B C^{2}$

 

$\therefore 289=225+B C^{2}$

Subtracting 225 from both the sides of the equation we get,

$\therefore 289-225=B C^{2}$

$\therefore 64=B C^{2}$

Let us take the square root we get,

$B C=\sqrt{64}$

$\therefore B C=8$

Therefore, the distance of the foot of the ladder from the building is $8 \mathrm{~m}$.