Question

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

Answer 1

`/_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!")`