Finding the Missing Angle with Given Cotangent

[Jason76]

There is no "Arcccotan", so how would you solve this?:

\(\displaystyle \sqrt{3} = cot(x)\)

Could you use some reciprocal version of \(\displaystyle \arctan\)?

 

[Deleted member 4993]

There is no "Arcccotan" function, so how would you solve this?:

\(\displaystyle \sqrt{3} = cot(x)\)

Could you use some reciprocal version of \(\displaystyle \arctan\)?

.... first convert it to TAN(x) = ? .... then use x = ARCTAN(?)

 

[Jason76]

.... first convert it to TAN(x) = ? .... then use x = ARCTAN(?)

\(\displaystyle \sqrt{3} = \cot(x)\)

\(\displaystyle \sqrt{3} = \dfrac{1}{\tan(x)}\)

\(\displaystyle \arctan(\sqrt{3}) = \arctan(\dfrac{1}{\tan(x)})\)

\(\displaystyle \dfrac{\pi}{3} = \dfrac{1}{x}\)

\(\displaystyle \dfrac{x\pi}{3} = 1\)

\(\displaystyle x\pi= 3\)

\(\displaystyle x = \dfrac{3}{\pi}\)

 

[Deleted member 4993]

There is no "Arcccotan", so how would you solve this?:

\(\displaystyle \sqrt{3} = cot(x)\)

Could you use some reciprocal version of \(\displaystyle \arctan\)?

\(\displaystyle \sqrt{3} = cot(x)\) → tan(x) = 1/√3 → x = π/6 ± k * π ...... (k = 0, 1, 2, 3....)

 

[soroban]

Hello, Jason76!

\(\displaystyle \text{Solve for }x\!:\;\cot x \,=\,\sqrt{3}\)

We have: .\(\displaystyle \cot x \:=\:\dfrac{\sqrt{3}}{1} \:=\:\dfrac{adj}{opp}\)

\(\displaystyle x\) is an angle in a right triangle with: \(\displaystyle adj = \sqrt{3},\: opp = 1\)
Pythagorus says: \(\displaystyle hyp = 2.\)

We have:

                        *
                     *  *
              2   *     *
               *        * 1
            *           *
         *   x          *
      *  *  *  *_ *  *  *
               √3

Look familiar?