For this case we must find the hypotenuse (H) or the BC side of the rectangular triangle shown in the figure.
We have to:
[tex]Tangent (B) = \frac {Cathet \ opposite} {Cathet \ adjacent}\\Tangent (45) = \frac {11} {BA}[/tex]
By clearing BA we have:
[tex]BA = \frac {11} {Tangent (45)}\\BA = \frac {11} {1}\\BA = 11[/tex]
The Pythagorean theorem, which states:
[tex]BC = \sqrt {(CA) ^ 2 + (BA) ^ 2}\\BC = \sqrt {(11) ^ 2 + (11) ^ 2}\\BC = \sqrt {(11) ^ 2 + (11) ^ 2}\\BC = \sqrt {2 * 11 ^ 2}\\BC = 11 \sqrt {2}[/tex]
Answer:
[tex]BC = 11 \sqrt {2}[/tex]
Option C
