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The number of sides of two regular polygons $A$ and $B$ are in the ratio $1 : 3$ . If each interior angle of polygon B is $168^{\circ}$ ; find each interior angle of polygon $A$ .
$144^{\circ}$
$134^{\circ}$
$124^{\circ}$
none of the above

A

124^{\circ}

B

none of the above

C

144^{\circ}

D

134^{\circ}

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Solution

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The number of sides of two regular polygons A and B are in the ratio $1 : 3$ . If each interior angle of polygon B is $168^{o}$ ;
Let number of sides of polygon A is $n_{1}$ and polygon B is $n_{2}$ .
So,
$\frac{n_{1}}{n_{2}} = \frac{1}{3}$ and $\frac{180^{o} (n_{2} - 2)}{n_{2}} = 168^{o}$
$=> n_{1} = \frac{n_{2}}{3}$ and $=> 180^{o} n_{2} - 168^{o} n_{2} = 360^{o}$
$=> n_{1} = \frac{n_{2}}{3}$ and $=> 12^{o} n_{2} = 360^{o}$
$=> n_{2} = 30$
Then,
$n_{1} = \frac{n_{2}}{3} = \frac{30}{3} = 10$
Each interior angle of polygon A = $\frac{180^{o} (n_{1} - 2)}{n_{1}} = \frac{180^{o} (10 - 2)}{10} = 144^{o}$

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