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# help

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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

#1
+262
+2
Apr 4, 2020
#2
+9142
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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
+112419
0

Very nice, hectictar   !!!

CPhill  Apr 4, 2020