Evaluate arccos(1/sqrt(3)), giving the result in both radians and degrees.

Step-by-step solution

1. 1

   Restate the notation as an equation. Asking for $\arccos(1/\sqrt3)$ means finding the unique angle $\theta$ in the principal range $[0,\pi]$ with

   $\cos\theta=\frac{1}{\sqrt{3}}$

   The range restriction is what makes the answer a single number rather than an infinite family.

2. 2

   Check the input against the special angles first. The cosines you are expected to recognise are

   $\cos 30^\circ=\frac{\sqrt3}{2}\approx0.8660,\quad \cos45^\circ=\frac{\sqrt2}{2}\approx0.7071,\quad \cos60^\circ=\frac12$

   Our input is $1/\sqrt3\approx0.57735$, which matches none of them, so no exact multiple of $\pi$ will do.

3. 3

   Locate the angle by bracketing. Since $0.5<0.57735<0.7071$ and cosine is decreasing on $[0,\pi]$, the angle must satisfy

   $45^\circ<\theta<60^\circ$

   This bracket is the cheap sanity check on whatever the calculator returns.

4. 4

   Evaluate numerically. To four decimal places,

   $\theta=\arccos(0.577350\ldots)=0.9553\text{ rad}$

   Converting with $180^\circ/\pi$:

   $\theta=0.9553166\times\frac{180}{\pi}=54.7356^\circ$

   which indeed lies inside the $45^\circ$–$60^\circ$ bracket.

5. 5

   Note where this angle actually comes from. $54.7356^\circ$ is the angle between a cube's space diagonal and an edge, and twice it, $109.47^\circ$, is the tetrahedral bond angle — so this particular arccosine shows up constantly in crystallography and chemistry.

6. 6

   State the answer both ways.

   $\arccos\left(\frac{1}{\sqrt3}\right)\approx0.9553\text{ rad}\approx54.74^\circ$