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Answer:
Solution given:
m∠ADB=(4x−12)°
m∠CDB=(3x+6)°
m∠ADC =?
Since diagonal BD bisect the angle <ADC
so
m∠ADB= m∠ADC
(4x-12)°=(3x+6)°
4x-3x=6-12
x=12+6
x=18°
again.
<ADB=m∠ADB+ m∠ADC=4×18-12+3×18+6=120°
So
<u>the m∠</u><u>ADC</u><u> </u><u>=</u><u>1</u><u>2</u><u>0</u><u>°</u>
