Can this Secant-Tangent Equation be Simplified Further?

[Cyclopse]

Homework Statement

Is it possible to simplify this even further?

Homework Equations

This is a Secant-tangent equation.
Secant: AD = 18/radical 3 AC = ??

The Attempt at a Solution

I cross multiplied to get 18=324radical 3 (AC)
and in the end i got 0 = (18radical3) (AC) which makes no sense

so is there a different way to simplify that?

 

[Mark44]

Homework Statement

Is it possible to simplify this even further?

Homework Equations

This is a Secant-tangent equation.
Secant: AD = 18/radical 3 AC = ??

The Attempt at a Solution

I cross multiplied to get 18=324radical 3 (AC)

Are you trying to solve for AC?

and in the end i got 0 = (18radical3) (AC) which makes no sense

When you divide both sides by 18, you will be left with 1 on the left side, not 0.

If you are trying to find AC, divide both sides by 324√3.

so is there a different way to simplify that?

 

[Cyclopse]

Are you trying to solve for AC?

Yes.
Here is the original question

 

[Mark44]

It doesn't seem to me that you have enough information.

How did you go from AB = 18 and AD = 18/√3 to 18 = 324√3 * AC?

 

[Cyclopse]

It doesn't seem to me that you have enough information.

How did you go from AB = 18 and AD = 18/√3 to 18 = 324√3 * AC?

I'm really terrible at maths...that's why I'm here.

 

[Mark44]

Draw a radius from O to B, and call its length r. The radius OB is perpendicular to AB, so that ABO is a right triangle.

Let θ = the angle at A.

Then, from your given information,

cos(θ) = 18/(r + 18/√3), and
sin(θ) = r/(r + 18/√3)

Now you have two equations in two unknowns, so you should be able to solve for both unknowns, although you only need r. Once you know r, it's pretty easy to get AC.

 

[Chestermiller]

Start out like Mark44 said by drawing a radius from O to B, and call its length r, but then apply the Pythagorean Theorem to the right triangle ABO to find r.

Chet

 

[Mark44]

Start out like Mark44 said by drawing a radius from O to B, and call its length r, but then apply the Pythagorean Theorem to the right triangle ABO to find r.

That's simpler than what I suggested. Simple is good!