Answer:
[tex]h=13\,\sqrt{3}[/tex]
[tex]x=52[/tex]
Step-by-step explanation:
Using the triangle formed in the far left, we can use the Pythagorean theorem to solve for h:
[tex]h^2+a^2=y^2\\h^2+39^2=(26\,\sqrt{3} )^2\\h^2=26^2\,(3)-39^2\\h^2=507\\h=\sqrt{507}\\h=13\,\sqrt{3}[/tex]
Now, using the right angle triangle on the right we solve for x:
[tex](x-a)^2+h^2=z^2\\(x-39)^2+507=26^2\\(x-39)^2=676-507\\(x-39)^2=169\\x-39=+/-13\\x=26\,\,\,or\,\,\,x=52[/tex]
since x has to be larger than "a" (39), we use the second answer:
[tex]x=52[/tex]
