Geometry

Here's an alternative method of solution.

Since the angle at B is a rt angle, it follows that B lies on the semi-circle for which AC is the diameter.

Then if we have, in a circle, two chords AC and BD intersecting at X,

AX.XC = BX.XD and since XD will equal BX, it follows that BX^2 = AX.XC = 3.12 = 36 etc. .

Tiggsy

Note higgsb, we don't have enough info  to  answer this

There's nothing wrong with the original problem, there is sufficient information, BX = 6.

With the altered problem, BX = 4sqrt(3).

Ah....guest is correct.....pardon my  stupid oversight....we can solve this with some trig....

We have  [ in the first situation]

tan ACB  =  BX / 12  = tan ABX  = 3 / BX  ...which implies that

BX / 12 = 3 /BX   so  

BX^2  = 3 * 12

BX^2  = 36

BX  = 6   !!!!!

In the second situation  AX  = 12   CX  = 4 .....as before

tan ACB  =  BX / 4  = tan ABX  = 12 / BX  ...which implies that

BX / 4 = 12 /BX   so  

BX^2  = 4 * 12

BX^2  = 48

BX  = √ [ 16 * 3 ]  =  4√3

Thanks, guest, for pointing out my error  ....!!!!