12. Find the missing angles of the quadrilateral ACBD in the given fig

1.	12. Find the missing angles of the quadrilateral ACBD in the given figure.
Answer» $\huge\underline\mathbb{\red S\pink {O}\purple {L} \blue {UT} \orange {I}\green {ON :}}$ In *Triangle* *ADC* By *ANGLE* *sum* *property* of *triangle* *40+70+ADC=180 ADC=180-110 ADC=70 Now* in *Triangle* *DBC* As *DB=BC So Angle* *BCD=BDC=x So in triangle* *DBC* by *angle* *sum* *property* of *triangle* 80+ x + x = 180 *2x=100 x=50 So BCD=*BDC=50 $\huge\underline\mathbb{\red F\pink {O}\purple {L} \blue {L} \orange {OW}\green {ME }}$

12. Find the missing angles of the quadrilateral ACBD in the given figure.

Answer»

$\huge\underline\mathbb{\red S\pink {O}\purple {L} \blue {UT} \orange {I}\green {ON :}}$

In Triangle ADC By ANGLE sum property of triangle

40+70+ADC=180

ADC=180-110

ADC=70

Now in Triangle DBC

As DB=BC

So Angle BCD=BDC=x

So in triangle DBC by angle sum property of triangle

80+ x + x = 180

2x=100

x=50

So BCD=BDC=50

$\huge\underline\mathbb{\red F\pink {O}\purple {L} \blue {L} \orange {OW}\green {ME }}$