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A regular heptagon and a regular 15-gon share a common edge $\overline{AB}$, as shown. Find angle BCD, in degrees.

 Dec 16, 2018
edited by Guest  Dec 16, 2018
edited by Guest  Dec 16, 2018
 #1
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An interior angle BAC of the the heptagon will measure  [(5/7)180] °

 

Draw  BC

And because BA = AC.....then in triangle ABC,  angle ABC = angle  ACB

 

And angle ACB  =  (180 - mBAC) / 2 =     ( 180 - (5/7)180 ) / 2   =  [ ( 2/7) 180 ] / 2  =  [180/7 ]°

 

Draw  BD......and the interior angle BAD of the 15-gon  =  [ (13/15)180 ]°

 

So angle DAC  =

 

360 - m BAD - m BAC   =   360 - [ 180 (5/7 + 13/15) ] °=  ( 360 - [ 180 (166/ 105)] )° =

 

[ 360 -   1992 / 7 ]°    =   [ 528 / 7 ]°

 

Draw DC

 

And in triangle  DAC....since AC = AD.....then angle ADC  = angle ACD

 

And angle ACD =    [ 180 - m DAC] / 2  =   [ 180 - 528/ 7 ] / 2   =

 

[732/ 14] °   =  [ 366/ 7]°

 

So......angle BCD  =  angle ACB + angle ACD = [ 180/ 7  + 366/ 7]°  =

 

[ 546/ 7]°   =  78°

 

 

cool cool cool

 Dec 16, 2018

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