# Thread: Question with special angles and Trig Identities

1. ## Question with special angles and Trig Identities

Hello again Everyone,

I had a question I was able to finally solve by myself but I was wondering if the following method is correct.

The problem is, $\displaystyle cot\theta = \frac{\sqrt{3}}{3}$

How I solved it...
$\displaystyle tan \theta = \frac{1}{cot \theta}$

$\displaystyle tan \theta = \frac{1}{\frac{\sqrt3}{3}}$ Which equals 1.732

Since I knew the question asked about special triangles I did the following
tan(30°) = 0.57735
tan(45°) = 1
tan(60°) = 1.73205
Therefore $\displaystyle \theta$ = 60°

Is this an acceptable method or the "right" method?

2. ## Re: Question with special angles and Trig Identities

Originally Posted by vaironxxrd
Hello again Everyone,

I had a question I was able to finally solve by myself but I was wondering if the following method is correct.

The problem is, $\displaystyle cot\theta = \frac{\sqrt{3}}{3}$

How I solved it...
$\displaystyle tan \theta = \frac{1}{cot \theta}$

$\displaystyle tan \theta = \frac{1}{\frac{\sqrt3}{3}}$ Which equals 1.732

Since I knew the question asked about special triangles I did the following
tan(30°) = 0.57735
tan(45°) = 1
tan(60°) = 1.73205
Therefore $\displaystyle \theta$ = 60°

Is this an acceptable method or the "right" method?
It is definitely a method. I wouldn't necessarily say it is the most direct or elegant method.

\displaystyle \displaystyle \begin{align*} \cot{\theta} &= \frac{\sqrt{3}}{3} \\ \cot{\theta} &= \frac{1}{\sqrt{3}} \\ \tan{\theta} &= \sqrt{3} \\ \theta &= 60^{\circ} + 180^{\circ} n \textrm{ where } n \in \mathbf{Z} \end{align*}

3. ## Re: Question with special angles and Trig Identities

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