<s>[INST] Why does my solutions manual take away or add 180 when finding $\theta$? These are the provided notes:



These are the provided questions:



I do not understand when I should subtract, add or leave the answer as is (The notes do not make sense to me very much). I do not, clearly, have an intuitive understanding of this. I hope someone can please please please show me thank you :)
 [/INST] Consider the following diagram:

The range of the arccosine function is $[0, \pi]$.  Since you are working in degrees, this corresponds to $[0^\circ, 180^\circ]$.  Thus, you can calculate the measure of the obtuse angle $\theta$ whose cosine is $-3/5$ directly since 
$$\cos\theta = -\frac{3}{5} \Longrightarrow \theta = \arccos\left(-\frac{3}{5}\right)$$ 
However, this will not work for sine or tangent since the range of the arcsine function is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ (or, in degrees, $[-90^\circ, 90^\circ]$) and the range of the arctangent function is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ (in degrees, $(-90^\circ, 90^\circ))$.
For an obtuse angle $\theta$ such that $\sin\theta = \frac{1}{4}$, we draw a right triangle in the second quadrant with opposite side of length $|y| = |1| = 1$, hypotenuse of length $r = 4$, and adjacent side of length $|x| = |-\sqrt{15}| = \sqrt{15}$.  The reference triangle we draw in the first-quadrant, has adjacent side of length $\sqrt{15}$, opposite side of length $1$, and hypotenuse of length $4$. The hypotenuse of the reference triangle forms an angle of $180^\circ - \theta$ with the positive $x$-axis, as shown in the diagram.  Since the sine function satisfies the property 
$$\sin(180^\circ - \theta) = \sin\theta$$
we obtain 
$$\sin(180^\circ - \theta) = \frac{1}{4}$$
Moreover, since $\theta$ is an obtuse angle, $180^\circ -\theta$ is an acute angle, so it falls within the range of the arcsine function. Thus,
$$180^\circ - \theta = \arcsin\left(\frac{1}{4}\right)$$
Solving for $\theta$ yields
$$\theta = 180^\circ - \arcsin\left(\frac{1}{4}\right)$$
It would be inconvenient to draw a right triangle for an obtuse angle $\theta$ such that $\sin\theta = 0.7890$.  However, we can still use the property that $\sin(180^\circ - \theta) = \sin\theta$ to conclude that 
$$\sin(180^\circ - \theta) = 0.7890$$
Since $180^\circ - \theta$ is an acute angle, we obtain
$$180^\circ - \theta = \arcsin(0.7890)$$
Now, solve for $\theta$.
For an obtuse angle $\theta$ such that $\tan\theta = -\frac{6}{11}$, we draw a second-quadrant angle with adjacent side of length $|x| = |-6| = 6$ and opposite side of length $|y| = |11| = 11$.  The hypotenuse has length $\sqrt{157}$.  The reference triangle in the first-quadrant has adjacent side of length $6$, opposite side of length $11$, and hypotenuse of length $\sqrt{157}$.  The angle the hypotenuse of the reference triangle forms with the positive $x$-axis is $180^\circ - \theta$.  Since $\theta$ is obtuse, $180^\circ - \theta$ is acute.  Since the tangent function has the property that 
$$\tan(180^\circ - \theta) = -\tan\theta$$
we obtain
$$\tan(180^\circ - \theta) = -\left(-\frac{6}{11}\right) = \frac{6}{11}$$
Since $180^\circ - \theta$ is an acute angle, it falls within the range of the arctangent function.  Thus,
$$180^\circ - \theta = \arctan\left(\frac{6}{11}\right)$$
Solving for $\theta$ yields
$$\theta = 180^\circ - \arctan\left(\frac{6}{11}\right)$$
For the obtuse angle $\theta$ such that $\tan\theta = -3.8522$, we use the property that $\tan(180^\circ - \theta) = -\tan\theta$ to conclude that 
$$\tan(180^\circ - \theta) = -(-3.8522) = 3.8522$$
Since $180^\circ - \theta$ is an acute angle,
$$180^\circ - \theta = \arctan(3.8522)$$
Now, solve for $\theta$.
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