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Question: If $\alpha$ is an angle in a triangle and $\tan{\alpha}=-2$, then one of the following is true:

a) $0<\alpha < \frac{\pi}{2}$

b) $\frac{\pi}{2}<\alpha < \pi$

c) Can't be decided.

d) There exist no such angle $\alpha$.

My reasoning was that there exist no such angle because of the following: Looking at a right triangle with an angle alpha and one of the sides 1,alpha should be positive between zero and 90 degrees (which is wrong).

enter image description here

Singe $1\cdot \tan{\alpha} = x,$ I don't see how a physical side on a triangle can be negative.

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    $\begingroup$ The triangle may be oblique (not a right triangle). $\endgroup$ – N. F. Taussig Apr 23 '17 at 10:09
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Without referring to the diagram, which is very misleading, the answer is (b), since:

(1) the angle must be between $\;0\;$ and $\;\pi\;$ radians as it belongs to a triangle (not written "a straight triangle" !), and

(2) It must such that the signs of sine and cosine as opposite, since

$$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}$$

and this only happens in the second and fourth quadrants.

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  • $\begingroup$ Seems my mind was so focused on a right triangle. Thanks a lot for the help! $\endgroup$ – Parseval Apr 23 '17 at 10:19
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  • Since $\tan \alpha$ is given a value (in this case, $-2$) it would contradict there is no angle $\alpha$ (d) and that we cannot determine the information (c).
  • If $\tan \alpha$ is negative, it cannot be in the first or third quadrant (all values are positive for $\tan \alpha$), $\therefore \alpha$ is negative in the second and fourth quadrants, which are $\frac {\pi}{2} < \alpha < \pi$ and $\frac {3 \pi}{2} < \alpha < 2 \pi$.

Thus b. is correct.

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