Answer
$(12,8)$
Work Step by Step
Step 1. Use the figure provided by the Exercise, use the distance formula with the given coordinates, we have:
\begin{cases} (x-22)^2+(y-32)^2=26^2 \\ (x-28)^2+(y-20)^2=20^2 \end{cases}
Step 2. Take the difference between the two equations and use the formula $a^2-b^2=(a-b)(a+b)$, we get:
$6(2x-50)-12(2y-52)=6\times46$ which gives $x-2y=-4$ or $x=2y-4$
Step 3. Substitute the above equation into one of the equations (2nd one used here):
$(2y-32)^2+(y-20)^2=20^2$ or $5y^2-168y+1024=0$
Step 4. The above equation can be factored as $(5y-128)(y-8)=0$ (you can also use the quadratic formula) and we get $y=8, 25.6$
Step 5. Discard $y=25.6$ as it will be far away from the Earth surface, we have $y=8$ and $x=12$
