Question
In a parallelogram ABCD, if $\angle\text{A}=(3\text{x}-20)^\circ,\angle\text{B}=(\text{y}+15)^\circ,\angle\text{C}=(\text{x}+40)^\circ,$ then find the values of x and y.

Answer

In parallelogram ABCD, $\angle\text{A}$ and $\angle\text{C}$ are opposite angles. We know that in a parallelogram, the opposite angles are equal. Therefore,$\angle\text{C}=\angle\text{A}$
We have $\angle\text{A}=(3\text{x}-20)^\circ$ and $\angle\text{C}=(\text{x}+40)^\circ$ Therefore, x + 40º = 3x - 20º x - 3x = -40º - 20º -2x = -60º x = 30º Therefore,$\angle\text{A}=(3\text{x}-20)^\circ$
$\angle\text{A}=[3(30)-20]^\circ$
$\angle\text{A}=70^\circ$
Similarly,$\angle\text{C}=70^\circ$
Also, $\angle\text{B}=(\text{y}+15)^\circ$ Therefore,$\angle\text{D}=\angle\text{B}$
$\angle\text{D}=(\text{y}+15)^\circ$
By angle sum property of a quadrilateral, we have:$\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}=360^\circ$
$70^\circ+(\text{y}+15)^\circ+70^\circ+(\text{y}+15)^\circ=360^\circ$
$140^\circ+2(\text{y}+15)^\circ=360^\circ$
$2(\text{y}+15)^\circ=360^\circ-140^\circ$
$2(\text{y}+15)^\circ=220^\circ$
$(\text{y}+15)^\circ=110^\circ$
$\text{y}=95^\circ$
Hence the required values for x and y are 30º and 95º respectively.

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