Question
In the given figure, compute the value of $x.$
✓
Answer
In the given figure, $\angle\text{DCB}=45^\circ,\angle\text{CBA}=35^\circ$ and $\angle\text{BAD}=35^\circ$
Here, we will produce $AD$ to meet $BC$ at $E$
Now, using angle sum property of the triangle In $\triangle\text{AEB}$
$\angle\text{BAE}+\angle\text{AEB}+\angle\text{EBA}=180^\circ$
$\angle\text{AEB}+35^\circ+45^\circ=180^\circ$
$\angle\text{AEB}+180^\circ=180^\circ$
$\angle\text{AEB}=180^\circ-80^\circ$
$\angle\text{AEB}=100^\circ$
Eurther, $BEC$ is a straight line.
So, using the property, "the angles forming a linear pair are supplementary",
we get $\angle\text{AEB}+\angle\text{AEC}=180^\circ$
$100^\circ+\angle\text{AEC}=180^\circ$
$\angle\text{AEC}=180^\circ-100^\circ$
$\angle\text{AEC}=80^\circ$ Also, using the property, " an exterior angle of a triangle is equal to the sum of its two opposite interior angles"
In $\triangle\text{DEC},\text{x}$ is its exterior angle
Thus, $\angle\text{x}=\angle\text{DCE}+\angle\text{DEC}$
$=50^\circ+80^\circ$
$=130^\circ$ Therefore, $\text{x}=130^\circ.$
Here, we will produce $AD$ to meet $BC$ at $E$
Now, using angle sum property of the triangle In $\triangle\text{AEB}$
$\angle\text{BAE}+\angle\text{AEB}+\angle\text{EBA}=180^\circ$
$\angle\text{AEB}+35^\circ+45^\circ=180^\circ$
$\angle\text{AEB}+180^\circ=180^\circ$
$\angle\text{AEB}=180^\circ-80^\circ$
$\angle\text{AEB}=100^\circ$
Eurther, $BEC$ is a straight line.
So, using the property, "the angles forming a linear pair are supplementary",
we get $\angle\text{AEB}+\angle\text{AEC}=180^\circ$
$100^\circ+\angle\text{AEC}=180^\circ$
$\angle\text{AEC}=180^\circ-100^\circ$
$\angle\text{AEC}=80^\circ$ Also, using the property, " an exterior angle of a triangle is equal to the sum of its two opposite interior angles"
In $\triangle\text{DEC},\text{x}$ is its exterior angle
Thus, $\angle\text{x}=\angle\text{DCE}+\angle\text{DEC}$
$=50^\circ+80^\circ$
$=130^\circ$ Therefore, $\text{x}=130^\circ.$
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