The sides of a quadrilateral are produced in order. What is the sum of the four exterior angles?
The sides of the quadrilateral $\mathrm{ABCD}$ are produced in order (according to figure).
Now, we need to find the sum of the exterior angles.
Since the angles made on the same side of straight line are $180^{\circ}$, i.e., linear pair, we have:
$\mathrm{a}+\mathrm{x}+\mathrm{b}+\mathrm{y}+\mathrm{c}+\mathrm{z}+\mathrm{w}+\mathrm{d}=180^{\circ}+180^{\circ}+180^{\circ}+180^{\circ}=720^{\circ}$
OR
Sum of the interior angles $+$ sum of exterior the angles $=180^{\circ} \times 4=720^{\circ}$
Since the sum of the interior angles of a quadrilateral is $360^{\circ}$, we have:
$w+x+y+z=360^{\circ}$
Substituing the value, we get:
$a+b+c+d=360^{\circ}$
$\therefore$ Sum of the exterior angles $=360^{\circ}$
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