I am a little confused about solving trigonometric functions with right triangles. I am given the problem: Let $ABC$ be a right triangle with $C = 90\,^{\circ}$ and sides opposite angles $A, B,$ and $C$ are denoted by $a, b,$ and $c$. Suppose $b=4$ and $sin A = \frac{5}{11}$. Evaluate $c$.
So I drew a right triangle with $c$ opposite of $90^\circ{}$ and I know that $sin$ is opposite/hypotenuse. So the triangle has hypotenuse = $11$, opposite = $5$, adjacent = $4$. Since I am evaluating for $c$, why is the answer ($c$) =$\frac{11\sqrt{6}}{6}$ instead of $11$? Am I missing something here?
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