#help
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All I could do
the two triangles have parallel sides as denoted by the line, thus the angles should be equal iirc?
so you can do 110+2x=180
So X = 35?
i believe so
Alright thanks
@jovial garnet has given 1 rep to @jovial root
no worries
How do you know they are parallel?
the little line
Doesnt that mean equal length
woops thats what i meant
not parallel
the angles should be the same tho
@jovial garnet sorry bout that, important correction to make
the lines are the same length not parallel
oh wait hold on ive been rather stupid
i do not believe so, my apologies
ive been tryna work it over
but ive only gotten this far
Thanks, I appreciate it🙏
im gonna keep with it till i find the answer
ill ping you if i find it, but see if that helps you
@jovial garnet has given 1 rep to @jovial root
Alright bet
looks cyclic (lol)
where did you get Q = x + 40? isn't Q = x + 20 by right triangles with perpendicular bisector of the vertical diagonal?
nvm my bad you are correct
i dont think so, the angle below x should be 80 then
its not exactly to scale, the supposedly equal lengths dont look exactly equal
but its pretty close
Yeah the shape isn't to scale
Or at least that's the intention of the question
I just end up proving x+z = 70
i have a diagram thats to scale
I have no idea how you get z or x
i think this can be done with a significant effort using trigonometry
then it will be an equation and we'll have to make a guess
and it will be a simple angle
It's definitely not meant to be that complicated
and then the other way to do it will be extremely slick, just drawing a few points
that you would absolutely not think of
at least, that's the way it was on a similar famous angle finding problem
This is similar to year 10 maths which is why I'm bewildered by the fact it's taking me so long to figure it out
there is in general no closed solution to such things
theres not a formula you can apply
Can you not setup simultaneous equations of some sort between the two triangles?
Though everhtime I try it I just prove x+z=70
we have to depend on the problem having some lucky angles that make things work out nicely
Because you get
2x+2z+40 =180
x+z+110=180
(x+40)+z+70=180
As far as I can take it
I'm wondering if there's maybe some law about triangles opposite to one another or similar that can be applied
Oh wait angles in a rhombus =360
Nvm that doesn't help you just prove x+z=70 again
what is coordbash
thats where you give every point a location on the x/y plane
usually
upon further reflection this is missing some information
otherwise it is unsolvable
Hold on a sec
Yeah I'm 85% sure it's unsolvable
how'd you get to it
you just cant get that with pure angle chasing
calculator
then its not the right answer
there is of course a better way
as i could draw you another rhombus with the same values that are given
which calculator did you use?
desmos, after i did the trig method i talked about, and geogebra also agrees
How did you get 20?
.
How can I do it without a calculator
i don't know
Alright
can't x also be 0?
but according to the diagram x can't be 0?
I don't think there's enough data for this to be solvable.
wat
i mean maybe x can be like greater than 90 or who knows
but you follow the diagram
this is the only value of x that looks anything like the diagram
x can't be over 90
but I get your point