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

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.

• We were given that $\sin(135)$. What does that mean? Feb 15, 2020 at 22:55
• Sorry, Martin, you're not making sense. "cos and tan is the radian" is a word salad, not a comprehensible mathematical statement. Feb 15, 2020 at 23:06
• In any event, it seems to me that not Martin but whoever put the question together is confused about reflex angles. Feb 15, 2020 at 23:06
• Writing $\sin=\frac{5}{13}$ is a bad thing to do. It's like writing $\sqrt\,=4$, or $+=11$. Remember "$\sin$" written by itself is just a sin. Feb 15, 2020 at 23:11
• The answers given for part (b) would make sense if the word reflex were replaced by the word obtuse. Feb 15, 2020 at 23:20

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 $$\tan\theta=\frac{5}{12}$$.

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 $$\theta = \arctan -\frac 5{12}$$.

• "Who gives a pile of fetid dingo kidneys"?? 😂 Feb 15, 2020 at 23:47