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Find two regular polygons such that the number of their sides may be as 3 to 4 and the number of degrees in an angle of the first to the number of grades in an angle of the second as 4 to 5.

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n1 = 6 and n2=8

n1 = -6 and n2=-8

n1 = 6 and n2=7

n1 = 8 and n2=6

answer is C.

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Detailed Solution

Let one side be

n_{1} = 3 x

and the other side be

n_{2} = 4 x \dots \dots \dots \dots \dots

... (i).
As we know that, number of polygon sides =

\frac{180^{\circ} (n - 2)}{n}

According to question,

\frac{\frac{180^{\circ} (n_{1} - 2)}{n_{1}}}{\frac{180^{\circ} (n_{2} - 2)}{n_{2}}} = \frac{\frac{180^{\circ} (3 x - 2)}{3 x}}{\frac{180^{\circ} (4 x - 2)}{4 x}}

\Rightarrow \frac{\frac{180^{\circ} (3 x - 2)}{3 x}}{\frac{180^{\circ} (4 x - 2)}{4 x}} = {(\frac{4}{4.5})}^{\circ} (given)

\Rightarrow \frac{180^{\circ} (3 x - 2)}{180^{\circ} (4 x - 2)} = {(\frac{4}{4.5} \times \frac{3}{4})}^{\circ}

\Rightarrow \frac{180^{\circ} \times 3 x - 180^{\circ} \times 2}{180^{\circ} \times 4 x - 180^{\circ} \times 2} = {(\frac{2}{3})}^{\circ}

\Rightarrow \frac{540^{\circ} x - 360^{\circ}}{720^{\circ} x - 360^{\circ}} = {(\frac{2}{3})}^{\circ}

\Rightarrow 3^{\circ} (540^{\circ} x - 360^{\circ}) = 2^{\circ} (720^{\circ} x - 360^{\circ})

\Rightarrow 1620^{\circ} x - 108 0^{\circ} = 1440^{\circ} x - 720^{\circ}

\Rightarrow 1620^{\circ} x - 1440^{\circ} x = 108 0^{\circ} - 720^{\circ}

\Rightarrow 180^{\circ} x = 360^{\circ}

\Rightarrow x = \frac{360^{\circ}}{180^{\circ}}

\Rightarrow x = 2

∴

one side be n1 = 6 and other side be n2=8.
Option 3 is correct.

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