Question
If the side $BC$ of $\triangle\text{ABC}$ is produced on both sides, then write the difference between the sum of the exterior angles so formed and $\angle\text{A}.$
✓
Answer
In the given problem, we need to find the difference between the sum of the exterior angles and $\angle\text{A}.$ 
Now, according to the exterior angle theorem $\text{ext}.\angle\text{C}=\angle\text{A}+\angle\text{B}\dots(1)$
Also, $\text{ext}.\angle\text{B}=\angle\text{A}+\angle\text{C}\dots(2)$
Further, adding $(1)$ and $(2)$ $\text{ext}.\angle\text{C}+\text{ext}.\angle\text{B}=\angle\text{A}+\angle\text{B}+\angle\text{A}+\angle\text{C}$
$=2\angle\text{A}+\angle\text{B}+\angle\text{C}\dots(3)$
Also, according to the angle sum property of the triangle,
we get, $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ\dots(4)$
Now, we need to find the difference between the sum of the exterior angles and $\angle\text{A}.$ Thus, $(\text{ext}.\angle\text{C}+\text{ext}.\angle\text{B})-\angle\text{A}=(2\angle\text{A}+\angle\text{B}+\angle\text{C})-\angle\text{A}$
$=\angle\text{A}+\angle\text{B}+\angle\text{C}$
$=180^\circ(\text{Using 4})$ Therefore, $(\text{ext}.\angle\text{C}+\text{ext}.\angle\text{B})-\angle\text{A}=180^\circ$
Now, according to the exterior angle theorem $\text{ext}.\angle\text{C}=\angle\text{A}+\angle\text{B}\dots(1)$
Also, $\text{ext}.\angle\text{B}=\angle\text{A}+\angle\text{C}\dots(2)$
Further, adding $(1)$ and $(2)$ $\text{ext}.\angle\text{C}+\text{ext}.\angle\text{B}=\angle\text{A}+\angle\text{B}+\angle\text{A}+\angle\text{C}$
$=2\angle\text{A}+\angle\text{B}+\angle\text{C}\dots(3)$
Also, according to the angle sum property of the triangle,
we get, $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ\dots(4)$
Now, we need to find the difference between the sum of the exterior angles and $\angle\text{A}.$ Thus, $(\text{ext}.\angle\text{C}+\text{ext}.\angle\text{B})-\angle\text{A}=(2\angle\text{A}+\angle\text{B}+\angle\text{C})-\angle\text{A}$
$=\angle\text{A}+\angle\text{B}+\angle\text{C}$
$=180^\circ(\text{Using 4})$ Therefore, $(\text{ext}.\angle\text{C}+\text{ext}.\angle\text{B})-\angle\text{A}=180^\circ$
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