[Closed] Maths problem...

10 deg?

x = 30

angles of a triangle all add up to 180.

I did it with a bit of trial and error but sure there will be a better way

If x = 30 then angles at E are, 30, 30 and 120.

this then makes angles at D 40, 100 & 40

The angles don't make sense. At the X in the lower part, the 50 is a larger angle than the 130...

[img] [/img]

why are the 50's in the middle larger than the 130's?

The angles don't make sense. At the X in the lower part, the 50 is a larger angle than the 130...

its not drawn correctly,

i assumed the 50/130 were written in the wrong angles...

It's not to scale, just a hand drawn copy which is why the physical look of the angles is weird.

It'll not be drawn to scale, to stop you measuring the angles.

There are various facts you can work out, using the sum of the angles in a triangle being 180deg and the sum in a quadrilateral being 360deg. Some rearranging and substituting should get you [i]x[/i].

i assumed the 50/130 were written in the wrong angles...

making the angles of the right triangle add up to 100?

Never assume as it makes and ass of u and me

Nope 20 deg

JoeG - Member

The angles don't make sense. At the X in the lower part, the 50 is a larger angle than the 130...

Angles are correct, originator can't draw worth a damn though...

Edit: and I'm sllloooowww 😀

20 degrees

I drew it out in CAD!!

How can we trust a question set by someone who can't even draw 10 triangles to scale?

*tuts*

Hmmm I can't find a way to quickly calculate x so I'm going to guess that you have to calculate all four unknown angles which will mean a system of 4 equations with 4 unknowns simultaneously.

20 by trial and error, but isn't this one of thse questions where you create two forumulae for x and the other unknowns and then use simultaneous equations/subtsituting the values of the two unknowns?

Thats what im doing at the moment, but I havent done this stuff for 20+ years.
Good to know I still can though.

Anyone solved it without brute-force/IT witchcraft? Just genuinely interested in an elegant solution. I probably still won't get it but hey.

And LOL at the protractor-wielding literalists

cmon bigyinn - one of us has to get the formula. I am even trying extending the lines outside the triangle.

60?

[img][url= http://farm8.staticflickr.com/7428/12088092873_fa7175d99a_o.jp g" target="_blank">http://farm8.staticflickr.com/7428/12088092873_fa7175d99a_o.jp g"/> [/img][/url]
[url= http://www.flickr.com/photos/55623703@N05/12088092873/ ]angle[/url] by [url= http://www.flickr.com/people/55623703@N05/ ]Adam Branston[/url], on Flickr[/img]

Well that's what it's supposed to look like anyway!

ads beat me to it.. CAD for the win! 😉

I thought it might have something to do with simultaneous equations, but I can't remember how to do them anymore!

Look forward to the answer and workings... for when this does the rounds on FacePoo. 😆

you have to use the 'bigger' triangles as well as the small ones - there's about 10ish triangles in there and you can use the larger ones to help calculate the angles on the smaller ones.

I make it 80 - I'm no mathematician though

I can't do maths without computer aids any more. Actually i never really could, just did enough to pass the right exams!!

Computer aids? that sounds nasty! 😯

So no one's solved it formulaicly? (sp)

If y is the upper angle at D, x = y-10. I guess you play with that a bit more?

I just substituted some unknown angle values using the (x+this+that)=180 in a couple of places - gave a formula but , meh

This is annoying me now. 20 seemed to work, but here is a formula attemp

CDB = 140
EDB = 130-x
CDE = 140-x

therefore, 130-x+140-x = 140

270-2x =140
-2x =140-270
-2x = -130
x= 65

but that looks wrong

I did it using known angles and substitutions and got x = 70, assuming the diagram is not to scale...

Are there lots of right answers?

x = 20 degrees?

EDIT: Put in other symbols for the other unkown angles (y, z, etc.), then as sum of all angles in each triangle = 180, re-arrange equations to the form of e.g. y = x + 50 + 30 and then put that into the sum for a triangle that you've used 'y' in and solve to find x? I imagine there's a clearer way of expressing that!

Come one I've refreshed enough times now. Someone put me out of my misery.
Edit - and before someone says - I know it's ****** 20!

Or, you can use the existing angles to give a set length to a particular side, and then calculate the relative lengths of the other sides, forming more triangle using 90 degrees where necessary, in order to calculate the angle that way. But that'd just be complicated 😀

Any triangle out of all the options that take in X works if X is out 20 and the angle at D above 40 (With me) has to be 110 to work in all options that use that angle.

The answer is indeed 20.

It's non-soluble without a length as the angle changes dependent on the height of the major triangle.

Assume the top triangle is
20
a b

and the complementary angles are c and x, so
20
a b
c x

we need to find a,b,c and x. Four equations and four unknowns;
a+b=160
a+x=150
b+c=140
x+c=130

Now you can solve these either as a matrix and take the inverse of the 4x4 or you can back substitute;

[160] [1100] [a]
[150] = [1001] [b]
[140] [0110] [c]
[130] [0011] [x]

Unfortunately the determinant of this matrix is zero, so there is no inverse. Hence no solution to the four equations. The problem is ill posed.

For back substitution;
c=130-x
b=140-c = 140-(130-x) = 10+x
a+b = a+(10+x) = 160 so a+x = 150

Hence you really only have three equations and four unknowns. That's why it is on Facebook 😉

Maths is more fun than bike fitting 😉

Cheers tired, that was becoming somewhat annoying!!!

Yep, Thanks Tired! I wasted about 15 min on that!

It can't be insoluble, can it ? After all, somebody managed to draw it.

Must be calculable using sin/cos etc I'd guess

(then again, I said 80 degrees)

You are welcome to try, but you will need a scale (length) before you bring out the Cos or Sin rules big guns 😈

60 degrees

It's non-soluble without a length as the angle changes dependent on the height of the major triangle.

You are right, but it is ill-posed. Unless anyone spots an error in my alebra. Didn't see one. four unknowns but only three independent equations. That means you can find x in terms of a ratio of another angle you can't find. So you'll have to draw it.

Ask yourself why it might be on Facebook 😉

Tired, if you think it is ill-posed, you find two solutions and draw them accurately 🙂

Never was much good at geometry, but I'll have a think about it. I think it must have a neat solution.

At airport waiting for a flight so this is perfect :))

@Tired's got it with the comment that the solution can be an equation, it doesn't have to be a single number ("42" and all that 🙂 )

Will double check once on plane where I can write out the various equations. @Tired you kissed the chance to accuse everyone else of going off at a tangent ... I'll get my coat