A regular heptagon and a regular 15-gon share a common edge $\overline{AB}$, as shown. Find angle BCD, in degrees.
+128474 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°
