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

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

Guest Jul 2, 2015

### Best Answer

#3
+18842
+10

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

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

heureka  Jul 3, 2015
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### 3+0 Answers

#1
+5

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

Guest Jul 2, 2015
#2
+81084
+10

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

CPhill  Jul 2, 2015
#3
+18842
+10
Best Answer

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

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

heureka  Jul 3, 2015

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