Answer
$cot~72^{\circ} = \frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}} = 0.325$
Work Step by Step
$cos~x = sin(90^{\circ}-x)$
We can find the value of $cos~72^{\circ}$:
$cos~72^{\circ} = sin(90^{\circ}-72^{\circ})$
$cos~72^{\circ} = sin(18^{\circ})$
$cos~72^{\circ} = \frac{\sqrt{5}-1}{4}$
$sin~72^{\circ} = sin~(90^{\circ}-18^{\circ}) = cos~18^{\circ}$
$cos^2~x+sin^2~x = 1$
$cos~x = \sqrt{1-sin^2~x}$
We can find the value of $cos~18^{\circ}$:
$cos~18^{\circ} = \sqrt{1-(sin~18^{\circ})^2}$
$cos~18^{\circ} = \sqrt{1-(\frac{\sqrt{5}-1}{4})^2}$
$cos~18^{\circ} = \sqrt{1-(\frac{6-2\sqrt{5}}{16})}$
$cos~18^{\circ} = \sqrt{\frac{10+2\sqrt{5}}{16}}$
$cos~18^{\circ} = \frac{\sqrt{10+2\sqrt{5}}}{4}$
Since $sin~72^{\circ} = cos~18^{\circ}$, then $sin~72^{\circ} = \frac{\sqrt{10+2\sqrt{5}}}{4}$
We can find the value of $cot~72^{\circ}$:
$cot~72^{\circ} = \frac{cos~72^{\circ}}{sin~72^{\circ}}$
$cot~72^{\circ} = \frac{\frac{\sqrt{5}-1}{4}}{\frac{\sqrt{10+2\sqrt{5}}}{4}}$
$cot~72^{\circ} = \frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}$
$cot~72^{\circ} = 0.325$