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PLEASE PLEASE HELP ASAPIn parallelogram ABCD, AD=19, EC=15, mABC=66°, mDAC=78° and mBDC=19°.Find mABD.

PLEASE PLEASE HELP ASAPIn parallelogram ABCD, AD=19, EC=15, mABC=66°, mDAC=78° and-example-1
User Kirk Powell
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1 Answer

20 votes
20 votes

we can see that

We know that, in a parallelogram the opposite angles are equal to each other, this means that


66=19+ADB

hence, we have that


\begin{gathered} \text{ADB}=66-19 \\ \text{ADB}=47 \end{gathered}

Now, from the figure, we can see that, in the triangle ADB, we know angle ADB=47 and angle DAB=78.

Since all interior angles of a triangle add up 180, we have


47+78+\text{ABD}=180

and from this equality, we can find angle ABD. It yields


\begin{gathered} 125+\text{ABD}=180 \\ \text{ABD}=180-125 \\ \text{ABD}=55 \end{gathered}

hence, the answer is ABD=55 degrees

PLEASE PLEASE HELP ASAPIn parallelogram ABCD, AD=19, EC=15, mABC=66°, mDAC=78° and-example-1
User Cody Brimhall
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