Isn't reflex angle between $180^\circ$ and $360^\circ$?

Question

I was solving this question.

Given that $\sin\theta = 5/13 $, find $\theta$, $\cos\theta$, and $\tan\theta$ where $\theta$

(a) is an acute angle, and

(b) is a reflex angle.

I could get the acute angle which is $22.62^\circ$. So, I did $360^\circ - 22.62^\circ$, to get the reflex angle, but the answer is $157.38^\circ$ which is $180^\circ - 22.62^\circ$. Why is that? Do I have a wrong understanding of a reflex angle? Please help.

Answer 1

It sounds like what you are trying to ask is the following:

Given that $\sin\theta=\frac{5}{13}$ find $\theta$, $\cos\theta$, and $\tan\theta$, where (a) $\theta$ is an acute angle, and (b) $\theta$ is a reflex angle.

I've just looked up reflex angle, and the definition says that $\theta$ is a reflex angle iff $180^{\circ}<\theta<360^{\circ}$. It follows that if $\theta$ is a reflex angle, then $\sin\theta<0$, so there is no solution if $\theta$ is a reflex angle.

If $\theta$ is acute than $\theta=\sin^{-1}\left(\frac{5}{13}\right)\approx22.6^{\circ}$.

Also, if $\theta$ is acute then $\cos\theta>0$ and $\tan\theta>0$. From the fact that $\sin^2\theta+\cos^2\theta=1$, you get that $\cos\theta=\frac{12}{13}$, which gives you that $\tan\theta=\frac{5}{12}$.

Answer 2

But $\sin (360 - \theta) \ne \sin\theta$.

so you can not figure that if $\sin \theta=\frac 5{13};0\le \theta < 360^{\circ}$, that that would mean the reflexive angle $\psi = 360-\theta$ will have $\sin \psi = \sin \theta$. In fact just the opposite; $\sin \psi = -\sin \theta$.

So:

$\sin \theta = \frac 5{13}$.

Find $\theta$: $\frac 5{13}>$ so $0< \theta < 180$. So if a) $\theta$ is accute then $\theta = \arcsin \frac 5{13}$. If b) $\theta$ is reflexive this is not possible and there is no solution.

$\cos \theta = \pm \sqrt{1-\sin^2 \theta} = \pm\frac {12}{13}$. So if a) $\theta$ is accute then $\cos \theta \ge 0$ and $\cos \theta = \frac {12}{13}$. If b) then $\theta$ is impossible so there is no solution.

$\tan \theta = \frac {\sin \theta}{\cos \theta}$. So if a) $\theta$ is accute then $\tan \theta = \frac {\frac 5{13}}{\frac {12}{13}} = \frac 5{13}$ and, again, for (b) there is no solution.

Now it's not sure if I'm supposed to leave it as $\theta = \arcsin \frac {5}{13}$ or if I'm supposed to do some fancy geometry and trigonometry to find out what the angle of right triangle with sides of $5,12,13$ is. But I'm not going to grub in the mud unless I know there's a potato down there.

I could punch keys into a calculator to find $\theta = \arcsin \frac {5}{13}$ but nobody wants some stupid decimal when $\arcsin \frac {5}{13}$ is much more descriptive than $\theta = 22.6whogivesapileoffetiddingokidneys1087668.....$

So $\theta = \arcsin \frac 5{13}=\arctan \frac 5{12}$. Final answer.

.....

I suspect the person asking the question meant obtuse.

In which case b) $90 < \theta < 180$ and as $\sin{180 - \theta}=\sin \theta$ we have $\theta = 180 -\arcsin \theta$. And $\cos \theta < 0$ so $\cos \theta =-\frac 5{12}$ and $\tan \theta =-\frac 5{12}$. By the way we can express $\theta = \arctan -\frac 5{12}$.