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.

  • 1
    $\begingroup$ The triangle may be oblique (not a right triangle). $\endgroup$ – N. F. Taussig Apr 23 '17 at 10:09

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


and this only happens in the second and fourth quadrants.

  • $\begingroup$ Seems my mind was so focused on a right triangle. Thanks a lot for the help! $\endgroup$ – Parseval Apr 23 '17 at 10:19
  • 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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