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Let \(A_1 A_2 ... A_{16}\) be a regular 16-gon.  Find angle \(A_7 A_9 A_{13}\).

 

 Feb 20, 2020

Best Answer 

 #1
avatar+24388 
+3

Let \(A_1 A_2 \ldots A_{16}\) be a regular 16-gon.  
Find angle \(A_7 A_9 A_{13}\).

 

\(\begin{array}{|lrcll|} \hline & 2\beta+ 4*\dfrac{360^\circ}{16} &=& 180^\circ \qquad (1) \\\\ & 2\alpha+ 2*\dfrac{360^\circ}{16} &=& 180^\circ \qquad (2) \\\\ \hline \\ (1)+(2): & 2\beta+ 4*\dfrac{360^\circ}{16} + 2\alpha+ 2*\dfrac{360^\circ}{16} &=& 180^\circ + 180^\circ \\\\ & 2\beta+ 2\alpha + 6*\dfrac{360^\circ}{16} &=& 360^\circ \\\\ & 2\beta+ 2\alpha + 135^\circ &=& 360^\circ \\\\ & 2\beta+ 2\alpha &=& 360^\circ - 135^\circ \\\\ & 2\beta+ 2\alpha &=& 225^\circ \\\\ & 2(\beta+ \alpha) &=& 225^\circ \\\\ & \beta+ \alpha &=& \dfrac{225^\circ}{2} \\\\ & \mathbf{ \beta+ \alpha} &=& \mathbf{112.5^\circ} \\\\ & \theta &=& \beta+ \alpha \\ & \mathbf{ \theta} &=& \mathbf{112.5^\circ} \\ \hline \end{array}\)

 

\(\angle A_7 A_9 A_{13} = \mathbf{112.5^\circ}\)

 

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 Feb 20, 2020
 #1
avatar+24388 
+3
Best Answer

Let \(A_1 A_2 \ldots A_{16}\) be a regular 16-gon.  
Find angle \(A_7 A_9 A_{13}\).

 

\(\begin{array}{|lrcll|} \hline & 2\beta+ 4*\dfrac{360^\circ}{16} &=& 180^\circ \qquad (1) \\\\ & 2\alpha+ 2*\dfrac{360^\circ}{16} &=& 180^\circ \qquad (2) \\\\ \hline \\ (1)+(2): & 2\beta+ 4*\dfrac{360^\circ}{16} + 2\alpha+ 2*\dfrac{360^\circ}{16} &=& 180^\circ + 180^\circ \\\\ & 2\beta+ 2\alpha + 6*\dfrac{360^\circ}{16} &=& 360^\circ \\\\ & 2\beta+ 2\alpha + 135^\circ &=& 360^\circ \\\\ & 2\beta+ 2\alpha &=& 360^\circ - 135^\circ \\\\ & 2\beta+ 2\alpha &=& 225^\circ \\\\ & 2(\beta+ \alpha) &=& 225^\circ \\\\ & \beta+ \alpha &=& \dfrac{225^\circ}{2} \\\\ & \mathbf{ \beta+ \alpha} &=& \mathbf{112.5^\circ} \\\\ & \theta &=& \beta+ \alpha \\ & \mathbf{ \theta} &=& \mathbf{112.5^\circ} \\ \hline \end{array}\)

 

\(\angle A_7 A_9 A_{13} = \mathbf{112.5^\circ}\)

 

laugh

heureka Feb 20, 2020
 #2
avatar+109450 
+1

Very crafty, heureka   !!!!

 

 

cool cool cool

CPhill  Feb 20, 2020
 #3
avatar+24388 
+2

Thank you, CPhill !

 

laugh

 Feb 20, 2020

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