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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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Just as a hint, the opposite and hypotheneuse sides are not 5 and 11 respectively. Rather, $\sin A = \frac5{11}$ means that the lengths of the sides of the opposite and hypotheneuse sides are in the ratio of $5:11$

For another hint, if the opposite side is $x$, then the hypotheneuse is $\frac{11}{5}x$, and remember your pythagorean theorem of course! $a^2 + b^2 = c^2$

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  • $\begingroup$ Why is Sine defined as opposite over hypotenuse? How do I know when it is a ratio of the lengths rather than the actual sides? $\endgroup$ – Kot Nov 7 '12 at 23:43
  • $\begingroup$ @StevenN You know it is a ratio because as you said, $\sin A = \frac{5}{11}$ and a fraction is a ratio. As for your first question, that is a lot of detail to go into. You should look up information on the unit circle in relations to trignometry. $\endgroup$ – Sidd Singal Nov 7 '12 at 23:49

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