# Construction of 145 degree angle

I've tried doing it but I end up only constructing 135 degree angle.I have to use ruler without divisions and compass.It must be done with system of isosceles and equilateral triangle and their properties ,e.g external angle and etc. and the bisector.

Can you give me directions? Thank You in advance!

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This is 29 72nths of a circle, eulers totient of 72 is 12 = $2^3 3$ that $3$ suggests that you need to be able to trisect an angle to construct this angle. – quanta Apr 20 '11 at 11:25
What does it mean to construct an angle "with isosceles and equilateral triangle and the bisector"? – Chris Eagle Apr 20 '11 at 11:28
Thank You but I'm 7th grade and we don't learn these things.We've only learned for construncting a isosceles and equilateral triangle.With the equilateral triangle I can construct 60 degree angle ,with it's external angle - 120 degree angle and with that 120 degree angle + bisector of the 60 degree angle I construct 150 degree angle but 145? – lam3r4370 Apr 20 '11 at 11:32
@lam3r4370: Equilateral triangles and bisectors can be constructed using the usual constructions (often referred to as "straightedge and compass" or "ruler and compass", as in lhf's comment), so the only thing in your description that might allow you to construct $145°$ is the "isosceles triangle" part -- unfortunately that's also the most unclear part of your description, so I think you should first clarify that. – joriki Apr 20 '11 at 12:24
A general remark that applies to several of the comments and answers: A lot of effort was expended trying to tell a 7th-grader what is and isn't possible with ruler and compass, which is something that us mathematicians are interested in for historical and theoretical reasons, and comparatively little effort was expended to first establish whether that had anything to do with what she or he was actually trying to do. – joriki Apr 20 '11 at 13:33

You cannot do it with an umarked ruler and compass.

Here is a way with a markable ruler and compass.

1. construct a 30 degree angle (bisect an equilateral triangle) and call the vertex $O$
2. draw a circle with centre $O$ and call where it meets the sides of the angle $A$ and $B$
3. mark the ruler with the radius of the circle
4. extend the line segment $AO$ beyond $O$, calling where it meets the circle again $C$, and then extend the line further beyond $C$
5. place the ruler so that it touches $B$, and so that it cuts the circle again and the extended line the distance apart marked on the ruler, then draw the line and calling these points $D$ and $E$ respectively
6. use what you know about angles and the isosceles triangles $BOD$ and $ODE$ to find the angle $OED$ (10 degrees)
7. add a right angle (90 degrees) and half a right angle (45 degrees) at E to get a total angle of 145 degrees ($10+90+45$)
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You cannot do it with an umarked ruler and compass. ,so maybe the math problem condition is incorrect.I think that they mean 135 degree angle ,which is very easy for me.Thank You! – lam3r4370 Apr 20 '11 at 15:39

If you can construct a 145 degree angle then you can construct a 55 degree angle by removing 90 degrees, and so a 10 degree angle, by removing 45 degrees. It is a classical theorem that a 10 degree angle cannot be constructed with ruler and compass. See http://en.wikipedia.org/wiki/Angle_trisection#Angles_may_not_in_general_be_trisected

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It's not entirely clear yet what constructions the OP is allowed to use -- it seems you're just assuming that it's ruler and compass; there's nothing about that in the question. – joriki Apr 20 '11 at 12:35
So yes, allowing trisections... you could do a neusis construction or use a carpenter's square, or a number of other devices... – J. M. Apr 20 '11 at 12:51

At the level of elementary geometry, I'd guess that it was a typo and that you are expected to compute a 135 degree angle. It's not immediately obvious, but can be figured out pretty easily (as you have already done) once you figure out what 135 degrees really looks like.

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did you read the first sentence to the question? :p – Willie Wong Apr 20 '11 at 18:07