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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
avatar+18712 
+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
avatar
+5

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

Guest Jul 2, 2015
 #2
avatar+78577 
+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
avatar+18712 
+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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