Find the hyperbolic distance in the upper hyperbolic plane 
Let $A=(0,112), B=(0,126), C=(98,112)$ be points in the hyperbolic upper half plane H. Find the hyperbolic distances $d_h(A,B), d_h(A,C), d_h(B,C)$. Every answer should be in the form of a logarithm.

So I worked out just $BC$ since I am sure $AB$ and $AC$ work out the same procedure.
To find $BC$:
we know $B=(0,126)$ and $C=(98,112)$
I found the midpoint=$(49,119)$ and I also found the reciprocal slope=$7$. Using this information, I found the x intercept to be $x=(32,0)$
Using the distance formula, I found CX=BX=130
and since the hyperbolic distance=$\ln\frac{\csc(B)-\cot(B)}{\csc(A)-\cot(A)}$, I obtained:
$$\ln\frac{\frac{130}{126}-\frac{-32}{126}}{\frac{130}{112}-\frac{-32}{112}}$$
I just want anyone to confirm if this looks right.
EDIT: 
From above:
$B=(X_1,Y_1)=(0,126)$
$C=(X_2,Y_2)=(98,112)$
$X=(X_3,Y_3)=(32,0)$
$CX=BX=130$
$$D(B,C)=\ln\frac{\frac{B_X}{Y_1}-\frac{X_1-X_3}{Y_1}}{\frac{CX}{Y_2}-\frac{X_2-X_3}{Y_2}}$$
Plugging in the values:
$$D(B,C)=\ln\frac{\frac{130}{126}-\frac{-32}{126}}{\frac{130}{112}-\frac{66}{112}}$$
 A: Not sure about your formula at all, and your "midpoint" is not on the hyperbolic line $BC$ (remember the hyperbolic line is a half circle)
https://en.wikipedia.org/wiki/Poincar%C3%A9_half-plane_model gives another formula:
$$ \operatorname{dist} (\langle x_1, y_1 \rangle, \langle x_2, y_2 \rangle) = \operatorname{arcosh} \left( 1 + \frac{ {(x_2 - x_1)}^2 + {(y_2 - y_1)}^2 }{ 2 y_1 y_2 } \right) $$
then according to https://en.wikipedia.org/wiki/Hyperbolic_function#Inverse_functions_as_logarithms
$$ \operatorname {arcosh} (x) = \ln \left(x + \sqrt{x^{2} - 1} \right) $$
so if you get the same answer via this method there must be something right about it.
if you get a different answer there is something wrong somewhere.
Good luck
Added later:
I did some calculating and I think your formula was wrong, you should multiply the values not divide them, but maybe nicer is to formulate them as:
$$ \ln \left( \frac{(1+ \cos(A) ) (1+ \cos(B) )}{\sin(A)\sin(B)} \right) $$ 
Or even more nicely as 
$$ \ln \left( \frac{1+ \cos(A) )}{\sin(A)} \right) + 
   \ln \left( \frac{1+ \cos(B) )}{\sin(B)} \right) $$ 
being the sum of the distances A - top of the line  and B - top of the line.
