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

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Solution

Sum of exterior angles of hexagon formed by producing sides of order $= 360^{o}$
$∴ (6 x - 1)^{o} + (10 x + 2)^{o} + (8 x + 2)^{o} + (9 x - 3)^{o} + (5 x + 4)^{o} + (12 x + 6)^{o} = 360^{o}$
$50 x + 10^{o} = 360^{o}$
$50 x = 360^{o} - 10^{o}$
$50 x = 350$
$x = \frac{350}{50}$
$x = 7$
$∴$ Angles are $(6 x - 1)^{o} : (10 x + 2)^{o} : (8 x + 2)^{o}; (9 x - 3)^{o} (5 x + 4)^{o}$ and $(12 x + 6)^{o}$
i.e. $(6 \times 7 - 1)^{o}; (10 \times 7 + 2)^{o}; (8 \times 7 + 2)^{o}; (9 \times 7 - 3)^{o}; (5 \times 7 + 4)^{o}; (12 \times 7 + 6)^{o}$
$41^{o}; 72^{o}; 58^{o}; 60^{o}; 39^{o}$ and $90^{o}$

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If the sides of a triangle are produced in order, prove that the sum of the exterior angles so formed is equal to four right angles.

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