pls help

Ok, 13 units     

Here is the great circle of the sphere  (The equator if you like)

CA will be the longest possible distance, in a straight line from A to C

Method:

1) Use the cosine rul to find angle alpha (keep it exact)

2) Find DX (exact)

3) find XC Pythagoaras's theorem (exact)

4) CA = 2* CX

I have not an arc distance, it is the straight distance.)

If you want the longest possible arc distance then you can find it afterwards easily enough. 

Welcome to the forum,

but please next time proof read your question and get rid of meaningless dollar signs.

Points A,B and C are on a sphere whose radius is 3. 

If AB=BC=10 what is the longest possible value of AC?

No 2 points on the sphere can be more than 6 units apart. (in a straight line). 

If you are talking the arc distance then the greatest distance is half of  9pi   (half the circumference of the great circle)

4.5pi is approx 14.13 units.

I would think that the points could be at most 4.5pi  units apart.

This is just a way to compute the problem without sine/cosine, but the other way is faster.

To maximize the value of AC, we need to maximize angle ABC, given that it is smaller than 180 degrees. Since A, B, and C are on the sphere, the circumcircle of triangle ABC is a cross section of the sphere. The larger this circle, the greater angle ABC is. So, we want to maximize the size of the circle. Thus, we want the circumcircle to be a great circle of the sphere.

In the diagram below ADB is a 10-13-13 triangle. If we let H be the foot of the perpendicular from D to AB, we see that DHA forms a 5-12-13 right triangle. So, DH = 12 and the area of ADB is

(1/2)·DH · AB = (1/2)·12·10 = 60

We can also write the area of ADB as (DB·AX)/2. Since DB=13, we know that AX = 120/13. Thus AC = 2·AX = 240/13 

Ok just use the diagram Melody has because I can't figure out how to upload an image-

Just imagine that H exists.