Reallyyy need help!

[jaunty quail]

I barely understand this.

[jaunty quail]

help!! 😭😭

[scarlet smelt]

are u prepared to see the most awesome diagram

[frosty harness]

That's trigo.
Just make 2 triangles. First, find the height of the first triangle at 7 feet of the building, then the distance when the height is of x-4, x being the height from the first triangle-4. You can then find the distance. Basically use
$a^2 + b^2 = c^2$, c² being the hypotenuse, which is the length of the triangle, a is the distance from the wall and b is the height

[winged slate]

Tittom_123

[scarlet smelt]

a true depiction of artistic intellect

[frosty harness]

I already explained them...

@jaunty quail

[scarlet smelt]

they might need a diagram

[frosty harness]

True

[jaunty quail]

my brain is melting rn

[frosty harness]

This might be a better graph tbh

[jaunty quail]

oh

[frosty harness]

Basically this

$AC = ED = 25$

$CD = x = \sqrt {25^2-7^2}$

$BD = x-4 = CD-4$

$BE = \sqrt {25^2-(x-4)^2}$

[winged slate]

Tittom_123

[frosty harness]

@jaunty quail, this is the best I can give you, I don't want to solve the problem for you, but I am pretty sure you can do it with that

[jaunty quail]

i solved 2 of them!

the 2*

i think im good no

now

[frosty harness]

Good. If you need more help, you can dm me, I will be happy to help you

[sly basin]

.close