Quadrilateral ABCD ​ is inscribed in circle O. What is ​ m∠A ​ ?

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Tony B Share
Mar 6, 2018

Answer:

#/_C=138^o#

#/_A = 180^o-138^o=42^o#

Explanation:

ABCD can be considered as two triangles back to back. The sum of the internal angles of a triangle is #180^o#. So the sum of the internal angles of ABCD #=2xx180^o=360^0#

Also #/_A+/_C=180^o=/_B+/_D#

#=>color(green)(2x+(3x-5))=180^0=color(brown)((x+5)+/_C)#

#color(green)(5x-5)=180#

Add 5 to both sides

#5x=185#

Divide both sides by 5

#x=185/5=37#
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#color(blue)("Consider "/_A+/_C)#

#(x+5)+/_C=180^o#

But #x=37# giving:

#37+5+/_C=180^o#

Subtract 42 from both sides (37+5=42)

#/_C=138^o#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check

#(x+5)+2x+(3x-5)+/_C=360#

#(37+5)+2(37)+[3(37)-5]+138=?#

#42+74+106+138=360 color(red)(larr" works!")#

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