Question

13. A ladder 25 m long reaches a window of a building 20 m above the ground. Determine the distance of the lader
from the building.
​

Answer 1

Answer -

Distance of the ladder from the building is 15 m.

Step-by-step explanation -

[ Refer the attachment for figure ]

Here -

• length of ladder = Hypoteneous = H = 25 m
• length of ladder from ground to window = Perpndicular = P = 20 m

We know that -

(Hypotenuse)² = (Perpendicular)² + (Base)²

or

(H)² = (P)² + (B)²

Substitute the known values above

$\implies\:\sf{(25)^2\: = \:(20)^2\:+ \:(B)^2}$

$\implies\:\sf{625\:=\:400\:+\:(B)^2}$

$\implies\:\sf{625\:-\:400\:=\:(B)^2}$

$\implies\:\sf{225\:=\:(B)^2}$

$\implies\:\sf{\sqrt{225}\:=\:B}$

$\implies\:\sf{B\:=\:15\:m}$

•°• Distance of ladder from the ground is 15 m.

Answer 2

Question :----- we have to find the distance of the ladder

from the building.

Given :-------

• length of ladder = AC = 25m
• Height of building = AB = 20m

Formula used :----

• Pythagoras theoram , (Hypotenuse)² = (Perpendicular)² + (Base)² = AB² +BC² = AC²
• Sin @ = Perpendicular / Hypotenuse
• Tan@ = Perpendicular /Base ...

we can solve it by :----

• Basic Pythagoras theoram .
• Trignometry Ratios ..

See First solution by Pythagoras theoram :-----

____________________________________________

$\pink{\bold{\underline{\underline{Solution(1)}}}}$

AC² = AB² +BC²

BC² = AC² - AB²

Putting values we get,

$BC^{2} = {(25)}^{2} - {(20)}^{2} \\ \\ BC^{2} = 625 - 400 \\ \\ BC^{2} = 225 \\ \\ BC^{2} = ( {15)}^{2} \\ \\ BC = 15 \: m$

So , the distance of the ladder from the building is 15m (Ans)

_____________________________

$\red{\bold{\underline{\underline{Solution(2)}}}}$

Lets solve it by Trignomteric Ratios Now, , ,

(Refer To image also for the value of @ = angle ACB)

we know,

sin@ = Perpendicular / Hypotenuse

Putting values we get,

$→ \sin( \theta) = \frac{20}{25} \\ \\→ \sin( \theta) = \frac{4}{5}$

Now we know that,

relationship of Tan@ in terms of sin@ is :---

$→ \tan( \theta) = \frac{ \sin( \theta)}{ \sqrt{1 - \sin( \theta)^{2} }}$

Putting values we get,

$→ \tan( \theta) = \frac{ \frac{4}{5} }{ \sqrt{1 - { (\frac{4}{5}) }^{2} } } \\ \\→ \tan( \theta) = \frac{ \frac{4}{5} }{ \frac{3}{5} } \\ \\ → \tan( \theta) = \frac{4}{ 3}$

$→ \tan( \theta) \: = \frac{4}{3} = \frac{p}{b} \\ \\ given \: p = 20 \\ \\ so \: multiply \: by \: 5 \: in \: num \: and \: \\ denominator \: we \: get \: = \\ \\ → \tan( \theta) = \frac{20}{15}$

Base = 15m

so, the distance of the ladder from the building is 15m ...

______________________________________

(Hope it helps you)

_____________________________

$\large\bold\star\underline\mathcal{Extra\:Brainly\:Knowledge:-}$

$\underline\textsf{Trignometric Ratios For Acute Angles :---}$

→ cosθ = sin(90-θ)

→ sinθ = cos(90-θ)

→ tanθ = cot(90-θ) = 1/cotθ

→ secθ = cosec(90-θ) = 1/cosθ

→ cosecθ = sec(90-θ) = 1/sinθ

→ cotθ = tan(90-θ) = 1/tanθ

$\mathcal{BE\:\:BRAINLY}$