Arcsin, arccos, and arctan

Consider a right-angled triangle, where we want to know the reference angle $\alpha$ , but all we know is the length ratio $\frac{O}{H}$ . For example, assume that $\frac{O}{H}=0.32$ . Can we find the angle $\alpha$ ? So, the question is, for what angle $\alpha$ is

\sin(\alpha)=0.32

Indeed, this is possible using the calculator. We need the function $\arcsin$ (say "arc sine") or $\sin^{-1}$ (say "inverse sine"). Depending on the calculator you will see one or the other notation. We then have

\sin(\alpha)=0.32 \rightarrow \alpha=\arcsin(0.32)=18.66^\circ

We can check if this is true by inserting $18.66^\circ$ into the sine: $\sin(18.66^\circ)=0.32$ .

Similarly,

to find $\alpha$ if we know the length ratio $A/H$ , we use $\arccos$ (say "arc cosine") or $\cos^{-1}$ (say "inverse cosine")
to find $\alpha$ if we know the length ratio $O/A$ we use $\arctan$ (say "arc tangent") or $\tan^{-1}$ (say "inverse tangent")

To summarise, we have

Definition 1

Finding the angle of given side length ratios:

\begin{array}{lll} \sin(\alpha) = \frac{O}{H}\text{ given} &\rightarrow& \alpha=\arcsin(\frac{O}{H})\\[4pt] \cos(\alpha) = \frac{A}{H}\text{ given} &\rightarrow& \alpha=\arccos(\frac{A}{H})\\[4pt] \tan(\alpha) = \frac{O}{A}\text{ given} &\rightarrow& \alpha=\arctan(\frac{O}{A}) \end{array}

Exercise 1

Determine the angle $\alpha$ in the triangles shown below.

Solution

(d) $\tan(\alpha)=\frac{3}{4}=0.75 \rightarrow \alpha=\arctan(0.75)=36.86...^\circ$
(e) $\cos(\alpha)=\frac{4}{5}=0.8 \rightarrow \alpha=\arccos(0.8)=36.86...^\circ$
(f) See the figure below. It is $h=10\cdot \sin(40^\circ)=6.428$ $u=10\cdot \cos(40^\circ)=7.66$ $v=20-u=12.34$ Thus, we have $\tan(\alpha)=\frac{h}{v}=0.52$ and therefore $\alpha=\arctan(0.52)=27.47^\circ$

Exercise 2

Determine all angles and side lengths.
A right-angled triangle has a hypotenuse that is three times longer than the smaller of the other two sides. Determine the exact angles.
Determine the angle of intersection between the graph of the function $f(x)=\frac{1}{2}x-1$ and the $x$ -axis.

Solution

$\frac{O}{A}=\frac{3.3}{5.8}=0.569 \rightarrow \alpha=\arctan(0.569)=\underline{29.64^\circ}$ . the other angle is $\beta=180^\circ-90^\circ-29.6^\circ=\underline{60.36^\circ}$ . The longest side can be determined using Pythagoras: $H=\sqrt{3.3^2+5.8^2}=\underline{6.67}$ .
$\frac{O}{H}=\frac{x}{3x}=\frac{1}{3} \rightarrow \arcsin(\frac{1}{3})=\underline{19.47^\circ}$ . The other angle different from $90^\circ$ is $90^\circ-19.47^\circ=\underline{70.53^\circ}$ .
We can select a right-angled triangle as shown in the figure below, and it follows $\frac{O}{A}=\frac{2}{4}=\frac{1}{2}\rightarrow \alpha=\arctan(\frac{1}{2})=\underline{26.565^\circ}$