n the figure below, the length of line segment CB is 58 units and the length of line segment BG is 120 units. What is the length of line segment GE? http://learn.flvs.net/webdav/assessment_images/educator_geometry_v14/pool_Geom_3641_0404_07/image0084e2ee688.gif 62√3 116 58√2 124 *I think the answer is C?*
I would think B...
look at the first triangle ABC - notice it is a right-angled triangle and one of the angles is 45, so what would the other angle be?
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
Vaguely, yes. I took geometry 3 years ago and got a B+ and A- but I don't remember most of it :/ I'm trying desperately to help a friend but some of this stuff I just don't remember. I know that tan = opposite ÷ adjacent cos= adjacent ÷ hypotenuse sin= opposite ÷ hypotenuse If I remember those correctly. I'm personally, currently in AlgII :/
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 \(62\sqrt{3}\)
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