Find x

Find x.

Do your own goddamn homework

I'm not doing your homework for you

16.97

Dude... are you blind? Its right there

x^2=16^2+16^2
x^2=512
x=16√2

>3 unknowns, three right-angled triangles

lrn2 system of equations

Please motivate first equation. How do you know the vertical line has length 16?

the right triangle is isosceles triangle so both its legs are equal.

12*sqrt(2)

thats wrong you retard...you cant assume that

Use the pytagorean thearom.

Im pretty sure you can.

She's probably off fucking some other guy. It's been over a year but I still miss her...

x = 20
guess and check.

There's no x, only this double mirrored c

How can x be 20 you retard, the bottom line is 18, x can't be bigger than 18.

Use the theory of relative height

Are you sure that the top angle is 90 degrees?

This is not solvable, we don't have enough information.

this

you need to know one of the remaining angles or another side

3,4,5

X=14.4

this

tan(a1)=2/y

tan(a2)=16/y

a1+a2=90o

tan(a1+a2)=(tan(a1)+tan(a2))/(1-tan(a1)*tan(a1))=inf

1-tan(a1)*tan(a1)=0

y^2=2*16

x=sqrt(16^2+y^2)=sqrt(16^2+2*16)

x=12*sqrt(2)

It's really not that hard.
Just use algebra and Pythagoras.

X has to be greater than 16.

Solved it for ya.

Name left unknown side y and middle unknown side z.

Pyt. theorem gives

x^2+y^2 =324 (large triangle)
x^2 -z^2=256 (left small triangle)
y^2-z^2=4 (right small triangle)

Subtract row 3 from row 2:
x^2-y^2 =252

add new eqn to row 1:
2*x^2=576

x^2=288

x=12*sqrt(2) roughly equal to

x=9

tan(a1)=2/y

tan(a2)=16/y

a1+a2=90o

tan(a1+a2)=(tan(a1)+tan(a2))/(1-tan(a1)*tan(a2))=inf

1-tan(a1)*tan(a2)=0

y^2=2*16

x=sqrt(16^2+y^2)=sqrt(16^2+2*16)

x=12*sqrt(2)

16 + 2 = 18
18 / 3 = 6
x = 6

wrong

lets call the other side (y), and the height (h)
then:

h^2 + 16^2 = x^2
h^2 + 2^2 = y^2
x^2 + y^2 = 18^2

h^2 = x^2 - 16^2
x^2 - 16^2 + 2^2 = y^2
x^2 + x^2 - 16^2 + 2^2 = 18^2
2 x^2 = 576
x^2 = 288
x = 16.97

Get over her, user!

this is correct

Lol!

all right

You fucking bag of shit

x != sqrt(162)

hope it helps

If the measures of it's angles are 90, 45, and 45 then the legs are equal

the hypotenuse is the sum of the angle of the square

Why am I a bag of shit?

shit ur right, sorry

At uni we don't even use pythagoras anymore so I totally forgot about him

h^2=x*y
h^2=32
h=4√2

4√2^2+16^2=x^2
32+256=x^2
x=√288
x=12√2