A circle centre O and radius 2cm is inscribed in an equilateral triangle ABC and touches the side BC at N. What is AN?

A circle centre O and radius 2cm is inscribed in an equilateral triangle ABC and touches the side BC at N. What is AN?

$$\\ \small{\text{when~ $ h = \overline{AN} $~ and one side is a}}\\\\ \qquad \tan{(60\ensurement{^{\circ}})} = \dfrac{h}{\frac{a}{2}} \qquad \tan{(30\ensurement{^{\circ}})} = \dfrac{r}{\frac{a}{2}} \\ \\ \dfrac{a}{2} = \dfrac{h}{ \tan{(60\ensurement{^{\circ}})} } = \dfrac{r}{ \tan{(30\ensurement{^{\circ}})} }\\\\\\ h= r\cdot \dfrac{ \tan{(60\ensurement{^{\circ}})} }{ \tan{(30\ensurement{^{\circ}})} } \small{\text{$ \qquad \tan{(60\ensurement{^{\circ}})} = \sqrt{3} \qquad \tan{(30\ensurement{^{\circ}})} = \dfrac{\sqrt{3} }{3} $}} \\\\\\ h= r\cdot \dfrac{ \sqrt{3}}{ \frac{\sqrt{3} }{3} }\\\\\\ h= 3\cdot r\\\\ h= 3\cdot 2~\text{cm}\\\\ h=6~\text{cm}\\\\ \mathbf{\overline{AN} = 6~\text{cm}}$$

$${\mathtt{AN}} = {\mathtt{6}}{cm}$$

Anonymous is correct....here's a pic.....

NB = 2√3   AB = 4√3  .......and by the Pythagorean Theorem.........

AN = √[(AB)^2 - (NB)^2]  =   √[(4√3)^2  -  (2√3)^2 ]  = √[48 - 12] = √36  = 6