The sides of a hexagon are produced in order. If the measures of exterior angles so obtained are $(6 x - 1)^{\circ}, (10 x + 2)^{\circ}, (8 x + 2)^{\circ}, (9 x - 3)^{\circ}, (5 x + 4)^{\circ}$ and $(12 x + 6)^{\circ};$ . Find each exterior angle.

41^{\circ}, 62^{\circ}, 58^{\circ}, 60^{\circ}, 39^{\circ}, 90^{\circ}

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41^{\circ}, 86^{\circ}, 56^{\circ}, 60^{\circ}, 39^{\circ}, 80^{\circ}

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41^{\circ}, 72^{\circ}, 58^{\circ}, 60^{\circ}, 39^{\circ}, 90^{\circ}

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41^{\circ}, 82^{\circ}, 60^{\circ}, 60^{\circ}, 36^{\circ}, 100^{\circ}

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Solution

The correct option is C $41^{\circ}, 72^{\circ}, 58^{\circ}, 60^{\circ}, 39^{\circ}, 90^{\circ}$
The sum of the exterior angles of any polygon is always equal to 360.
Exterior angles are $(6 x - 1)^{\circ}, (10 x + 2)^{\circ}, (8 x + 2)^{\circ}, (9 x - 3)^{\circ}, (5 x + 4)^{\circ} a n d (12 x + 6)^{\circ}$
Now,
$(6 x - 1)^{\circ} + (10 x + 2)^{\circ} + (8 x + 2)^{\circ} + (9 x - 3)^{\circ} + (5 x + 4)^{\circ} + (12 x + 6)^{\circ} = 360^{o}$
$=> (50 x + 10)^{o} = 360^{o}$
$=> x = 7$
Each Exterior angle
$=> (6 x - 1)^{o} = 6 \times 7 - 1 = 41^{o}$
$=> (10 x + 2)^{o} = 10 \times 7 + 2 = 72^{o}$
$=> (8 x + 2)^{o} = 8 \times 7 + 2 = 58^{o}$
$=> (9 x - 3)^{o} = 9 \times 7 - 3 = 60^{o}$
$=> (5 x + 4)^{o} = 5 \times 7 + 4 = 39^{o}$
$=> (12 x + 6)^{o} = 12 \times 7 + 6 = 90^{o}$

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