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\(\text{Let $B,$ $A,$ and $D$ be three consecutive vertices of a regular $18$-gon. A regular heptagon is constructed on $\overline{AB},$ with a vertex $C$ next to $A.$ Find $\angle BCD,$ in degrees. }\)

 Apr 4, 2020
 #2
avatar+9142 
+4

measure of an interior angle of a regular n-gon   =   ( n - 2 ) * 180°  /  n

 

m∠BAD   =   ( 18 - 2 ) * 180°  /  18   =   160°

 

m∠BAC   =   ( 7 - 2 ) * 180°  /  7   =   (900/7)°

 

Now let's try to find  m∠DAC

 

m∠DAC  +  m∠BAD  +  m∠BAC   =   360°

 

So...

 

m∠DAC   =   360°  -  m∠BAD  -  m∠BAC   =   360°  -  160°  -  (900/7)°   =   (500/7)°

 

Notice that  △BAC  is an isosceles triangle because  AB  =  AC

 

And also  △ACD  is an isosceles triangle because  AD  =  AC

 

The measure of each base angle of an isosceles triangle   =   ( 180° - measure of vertex angle ) / 2

 

m∠ACD   =   ( 180°  -  m∠DAC ) / 2   =   ( 180°  -  (500/7)° ) / 2   =   (380/7)°

 

m∠ACB   =   ( 180°  -  m∠BAC ) / 2   =   ( 180°  -  (900/7)° ) / 2   =   (180/7)°

 

Now we can find  m∠BCD

 

m∠BCD   =   m∠ACB  +  m∠ACD   =   (380/7)°  +  (180/7)°   =   80°

 Apr 4, 2020
 #3
avatar+112419 
0

Very nice, hectictar   !!!

 

cool cool cool

CPhill  Apr 4, 2020

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