Rhombus ABCD has a diagonal BD¯¯¯¯¯. m∠ADB=(4x−12)°

m∠CDB=(3x+6)°

What is the m∠ADC ?

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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

the m∠ADC =120°

Answer is in attachment hope it helps you...