# Geometry Problem

1. Feb 26, 2005

### Jameson

In $$\bigtriangleup ABC , AB = BC$$ and $$CD$$ bisects angle C.
If $$y = \frac{1}{3}x$$ then z = ....

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Ok. I've tried to make a bunch of substitutions for each of the angles and I can't seem to solve any for z.

I have found these two equations:

$$y + z + c = 180$$

and

$$x + z + \frac{c}{2} = 180$$

I know I can make substitutions/variants of these, but I still am stuck.
Any help is appreciated.

Jameson

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2. Feb 26, 2005

### cepheid

Staff Emeritus
Notice that angle CDB = 180 - x. So:

(180 - x) + y + C/2 = 180 (from the upper triangle).

(180 - x) + x/3 + C/2 = 180

C/2 = 2x/3

Can you proceed from here?

3. Feb 26, 2005

### Jameson

I got the same thing. $$\frac{c}{2} = \frac{2x}{3}$$ but unfortunately I still can't seem to do the right substitution.

Substituting (2x)/3 for c/2 I get the equation:

$$x + \frac{2x}{3} + z = 180$$

This still leaves two variables. I just can't see it for some reason.

4. Feb 26, 2005

5. Feb 26, 2005

### Jameson

This is from a published SAT book from College Board, so I think there is a solution.

6. Feb 26, 2005

### cepheid

Staff Emeritus
I, too, assumed that you were meant to solve for z in terms of x, since y was given in terms of x. I don't see what else you can do with it...

7. Feb 26, 2005

### apchemstudent

no no, see how AB = BC, That's a very important clue. We know triangle ABC is an isoceles triangle with angle A = angle C. You can come up with an equation to relate x directly to z and substitute it back into one of the equations to solve for z.

Angle C = 4/3 x = Angle A = z

you should be able to take it from here.

8. Feb 27, 2005

### honestrosewater

Eh, doesn't 2z + y = 180? So knowing y = x/3 you can solve for z in terms of x.

Plus, if it's the CB's "10 Real SAT's", it has the answers. And any other SAT prep book should have the answers too.

Last edited: Feb 27, 2005
9. Feb 27, 2005

### apchemstudent

ok, that equation is essential to solve for z. However, an actual value can be determined here. It's not impossible. I solved it and i got z = 80 degrees. Let me know if you want me to show my work.

10. Feb 27, 2005

### Jameson

apchemstudent... 80 degrees is the correct answer. I would be most gracious if you could show your work.

Jameson

11. Feb 27, 2005

### apchemstudent

ok, we set angle BDC = 180 - x, so that we can solve for Angle BCD

BCD + DBC + CDB = 180 degrees
BCD = 180 -1/3 x - 180 + x
BCD = 2/3 x

BCD = 1/2 BCA so BCA = 4/3 x

Again we know that it is an isoceles triangle with angle BCA = BAC
z = 4/3 x

so BAC + ABC + BCA = 180
z + 1/3 x + 4/3 x = 180
3x = 180
x = 60 <---- you could've substituted x in terms of z and solved for z but i find it easier this way.

now z = 4/3 (60) = 80 degrees.