In a pentagon, two angles are $40^{o} a n d 60^{o}$ , and the rest are in the ratio 1 : 3 : 7. Find the biggest angle of the pentagon.

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Solution

In a pentagon, two angles are $40^{o} a n d 60^{o}$

$∴$ Sum of remaining 3 angles $= 3 \times 180^{o} = 540^{o} - 40^{o} - 60^{o} = 540^{o} - 100^{o} = 440^{o}$

Ratio in these 3 angles = 1 : 3 : 7

Sum of ratios = 1 + 3 + 7 = 11

$∴$ Biggest angle $= \frac{440^{o} \times 7}{11} = 280^{o}$

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In a pentagon, two angles are 40o and 60o, and the rest are in the ratio 1 : 3 : 7. Find the biggest angle of the pentagon.

In a pentagon, two angles are 40o and 60o ∴ Sum of remaining 3 angles =3×180o=540o−40o−60o=540o−100o=440o Ratio in these 3 angles = 1 : 3 : 7 Sum of ratios = 1 + 3 + 7 = 11 ∴ Biggest angle =440o×711=280o

In a pentagon, two angles are $40^{o} a n d 60^{o}$ , and the rest are in the ratio 1 : 3 : 7. Find the biggest angle of the pentagon.

In a pentagon, two angles are $40^{o} a n d 60^{o}$

$∴$ Sum of remaining 3 angles $= 3 \times 180^{o} = 540^{o} - 40^{o} - 60^{o} = 540^{o} - 100^{o} = 440^{o}$

Ratio in these 3 angles = 1 : 3 : 7

Sum of ratios = 1 + 3 + 7 = 11

$∴$ Biggest angle $= \frac{440^{o} \times 7}{11} = 280^{o}$