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A regular heptagon and a regular 18-gon share a common edge AB, as shown. Find angle BCD, in degrees.

 

 

 Jan 10, 2020
 #1
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Angle BCD is 75 degrees.

 Jan 11, 2020
 #2
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Angle BCD = 80°, but don't ask me how I've done it.  indecision

 Jan 11, 2020
edited by Dragan  Jan 11, 2020
edited by Dragan  Jan 11, 2020
edited by Dragan  Jan 11, 2020
edited by Dragan  Jan 23, 2020
 #3
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Because of the 18-gon, angle BCD is 72 degrees.

 Jan 11, 2020
 #4
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A regular heptagon and a regular 18-gon share a common edge AB, as shown.

Find angle BCD, in degrees.

\(\begin{array}{|rcll|} \hline \mathbf{\angle 1} &=& \dfrac{(7-2)180^\circ}{7}\mathbf{=\dfrac{5*180^\circ}{7}} \\ \hline 2*\angle 2 + \angle 1 &=& 180^\circ \\ 2*\angle 2 &=& 180^\circ - \angle 1 \\ 2*\angle 2 &=& 180^\circ - \dfrac{5*180^\circ}{7} \\ 2*\angle 2 &=& 180^\circ \left(1 - \dfrac{5}{7} \right) \\ 2*\angle 2 &=& 180^\circ \left( \dfrac{2}{7} \right) \\ \mathbf{\angle 2} &=& \mathbf{\dfrac{180^\circ}{7}} \\ \hline \mathbf{\angle 3} &=& \dfrac{(18-2)180^\circ}{18} =\dfrac{16*180^\circ}{18} \mathbf{=160^\circ } \\ \hline \angle 4 &=& 180^\circ - \angle 3 \\ \angle 4 &=& 180^\circ - 160^\circ \\ \mathbf{\angle 4} &=& \mathbf{20^\circ} \\ \hline \mathbf{\angle 5} &=& 180^\circ - \angle 1 \mathbf{= 180^\circ \left( \dfrac{2}{7} \right)} \\ \hline 2*\angle 6 + \angle 4 + \angle 5 &=& 180^\circ \\ 2*\angle 6 + 20^\circ + 180^\circ \left( \dfrac{2}{7} \right) &=& 180^\circ \\ 2*\angle 6 + 180^\circ \left( \dfrac{2}{7} \right) &=& 160^\circ \\ 2*\angle 6 &=& 160^\circ - 180^\circ \left( \dfrac{2}{7} \right) \\ \mathbf{\angle 6} &=& \mathbf{80^\circ - \dfrac{180^\circ}{7}} \\ \hline \angle BCD &=& \angle 2 + \angle 6 \\ \angle BCD &=& \dfrac{180^\circ}{7}+ 80^\circ - \dfrac{180^\circ}{7} \\ \mathbf{\angle BCD} &=& \mathbf{80^\circ} \\ \hline \end{array}\)

 

 

laugh

 Jan 13, 2020

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