I really just dont know where to start on this one to be honest. I thought about making a triangle BEA or CBG but not sure how that would work. Thansk for the help.

Guest Feb 6, 2020

#1**+2 **

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Aha! CBD ~ CGA!

Both are 30-60-90.

1. Set CG = \(x\), and set CB = \(y\)

2. CD is \(x+1\)

3. Based on 30-60-90 properties, \(\frac{x+1}{2}=y\)

4. Write a similarity proportion

\(\frac{y}{x+1}=\frac{x}{y+13}\)

So now we have two equations...

Have fun solving!

CalculatorUser Feb 6, 2020

#2**+3 **

Here is my solution :

https://imgur.com/a/hhC4nVl

I just used some basic geometry theories like Equilateral Triangle, Pythago, etc.

Hope it can help :)

noobfromvn26 Feb 6, 2020

#4**+3 **

Triangles BCD, ACG, and GDN are similar; all of them are 30-60-90

Let's denote the intersection point of AF and BD with a leter N

BN = 13 * tan(30°) = 7.505

DN = 1 / cos (30°) = 1.155

BD = BN + DN = 8.66

CD = BD / cos (30°) = 10

Circle radius r = CD / 2 = 5

EG = sqrt [ r² - ( r - 1 )² ] = 3

CG = CD - GD = 9

AG = CG / tan (30°) = 15.588

EA = AG - EG = 12.588 _{}

Dragan Feb 6, 2020