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I would think B...

45º?

correct - so that implies it is also isoceles

which means AB=BC=58

Oh, they're both special triangles...one is 45-45-90, the other is 30-60-90

agreed?

Agreed.

so what would AG=

62? I think

yes

now use the fact that:\[\cos(60)=0.5\]to work out AE

sorry - use the fact that:\[\tan(60)=\sqrt(3)\]to work out GE

Well, I'm not really sure how to do that. Sorry, I'm dumb when it comes to math. :/

are you familiar with trigonometry (sin/cos/tan)?

62(60)sqrt3

if not, then this might help you: http://www.mathsisfun.com/algebra/trig-four-quadrants.html

good, so look at triangle AEG and note that:\[\tan(60)=\frac{GE}{AG}=\frac{GE}{62}\]

which implies:\[GE=62\times\tan(60)\]

does that make sense?

Ah! Yes. I think so. I had to figure out where the 62 came from again xD

good, now some angles have /well known/ values for sin/cos/tan

60 is one such angle where: \(\tan(60)=\sqrt{3}\)

can you work it out now?

Let me try, and then I'll post it :D

ok

107.26 is 62*1.73

roughly yes - you should leave it in terms of the radical, so the answer is \(62\sqrt{3}\)