# Thread: Find angle from cosec and tan.

1. ## Find angle from cosec and tan.

Given that cosec A=-2, tan A=√3÷3 and −π <A<π, find the exact value of the angle A in radians. Justify your answer.

Originally Posted by Gareth212
Given that cosec A=-2, tan A=√3÷3 and −π <A<π, find the exact value of the angle A in radians. Justify your answer.
You should know that the point $\displaystyle (x,y)$ is associated with angle $\displaystyle A$ where $\displaystyle x^2+y^2=r^2,~\csc(A)=\frac{r}{y},~\&~\tan(A)=\frac {y}{x}$.

This doesn't ring any bells. Isn't that (what you've said) with relation to a circle? I don't understand how you've come up with this? Super confused now!
Can you solve it? Show me how you see this from my question.
Any further response will be greatly appreciated.

if you know your unit circle as you should ...

$\displaystyle \csc{A} = -2 \implies \sin{A} = -\frac{1}{2}$

$\displaystyle A = -\frac{\pi}{6}$ or $\displaystyle A = -\frac{5\pi}{6}$

$\displaystyle \tan{A} = \frac{\sqrt{3}}{3}$

$\displaystyle A = \frac{\pi}{6}$ or $\displaystyle A = -\frac{5\pi}{6}$

so ... what's that tell you?