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I know how to find reference angles, but I'm having a hard time using reference angles to find an angle measure of interest.

For example, say we have an initial ray along the x-axis inside a unit circle and it meets the end of the circle at $P_1(4, 0)$, with its vertex of course at $(0, 0)$. We have the terminal ray passing counter clockwise at $P_2(-3.77, 1.34)$ on the circle. The angle measure here is $\alpha$. This is what we're looking for (rounded to one decimal).

I understand intuitively that $\alpha=\pi-\theta$, where $\theta$ is the angle measure of a constructed right triangle with height $1.34$ units, length $-3.77$ units, and hypotenuse $4$ units. But there is no way for me to complete this without already knowing what $\theta$ is. So it looks like I need to use a reference angle to find $\theta$, then use that to find $\alpha$.

First I tried using interior angles. One angle must be $90^\circ$, and so $\theta$ and its interior angle opposite to it (say $\beta$) must both add up to $90^\circ$. \begin{align*} \sin(\beta)=& -\frac{3.77}{4} \\ \arcsin(-\frac{3.77}{4})=& \beta \\ \beta =& -1.23 \ \text{or} \ 70.48^\circ \\ 70.48^\circ+\theta=&90^\circ \\ \theta =& 2.8 \ \text{rad} \\ \alpha =& \pi - 2.8 = 0.3 \ \text{rad} \end{align*} then I tried using an inverse function to find $\theta$ directly. \begin{align*} \tan(\theta)=&1.34/3.77 \\ \theta =& \arctan(-1.34/3.77) \\ \theta =&-0.341 \ \text{rad} \\ \alpha =& \pi-(-0.341) \\ \alpha =& 2.8 \ \text{rad}& \end{align*}

It seems 2.8 is the best answer I've gotten of all my attempts. Can anybody verify?

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There is no need for the negative signs, you are already considering the acute interior angles:

$\sin(\beta)= \dfrac{3.77}{4}$ and $\tan (\theta) = \dfrac{1.34}{3.77} \implies \theta = 0.341 $ rad

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  • $\begingroup$ Sorry, can you explain a little more why the acute interior angle has no need for a negative sign? $\endgroup$ – Lex_i Dec 7 '19 at 16:22
  • $\begingroup$ Why is the sine function having 3.77 on the numerator and tangent has 1.34? They should both be 1.34 $\endgroup$ – Andrei Dec 7 '19 at 16:23
  • $\begingroup$ @Andrei $\beta$ is different from $\theta$ $\endgroup$ – ab123 Dec 7 '19 at 16:26
  • $\begingroup$ @Lex_i $\sin$ and $\tan$ take positive values for angles in the range $(0, 90^\circ)$ $\endgroup$ – ab123 Dec 7 '19 at 16:27
  • $\begingroup$ @ab123 oops. sorry. I did not notice that you are looking at the complement. my appologies $\endgroup$ – Andrei Dec 7 '19 at 16:30

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