KelbyT419763 KelbyT419763
  • 25-10-2022
  • Mathematics
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PLEASE PLEASE HELP ASAPIn parallelogram ABCD, AD=19, EC=15, mABC=66°, mDAC=78° and mBDC=19°.Find mDAB.

PLEASE PLEASE HELP ASAPIn parallelogram ABCD AD19 EC15 mABC66 mDAC78 and mBDC19Find mDAB class=

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HarmonZ295281 HarmonZ295281
  • 25-10-2022

the measures of the opposite angles of a parallelogram are equal

so

[tex]\text{ADC}=\text{ABC}[/tex]

and

[tex]DAB=\text{BCD}[/tex]

the sum of the total angles of a parallelogram is 360

so

[tex]\text{ADC}+\text{ABC}+\text{DAB+BCD=}360[/tex]

where ABC=ADC=66

and

BCD=DAB

so replacing

[tex]\begin{gathered} 66+66+\text{DAB}+\text{DAB}=360 \\ 132+2\text{DAB}=360 \end{gathered}[/tex]

now solve DAB

[tex]\begin{gathered} 2\text{DAB}=360-132 \\ \text{DAB}=\frac{228}{2} \\ \text{DAB}=114 \end{gathered}[/tex]

the measure of DAB is 114°

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