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Thanks!

 Oct 13, 2019
 #1
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Hello Guest!

 

\(\angle ACD=\angle ABC =x\\ \angle BDC=\angle BCD=y \)

 

so:

2y+x=180°

x=180°-2y

x=180°-2y

48°+(180°-y)+x=180°

48°+(180°-y)+(180°-2y)=180°

-3y=(180-180-180-48)°

-3y=-228°

y=76°

x=180°-2*76°

x=28°

\(\angle ABC=28°\)

laugh  !

 Oct 13, 2019
edited by asinus  Oct 13, 2019

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