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

According to the properties of a polygon, if a polygon has n sides, then the sum of its interior angles is (2n - 4) x 90°.

For a pentagon with 5 sides: n = 5

The sum of its interior angles is:

(2 x 5 - 4) x 90°

= (10 - 4) x 90°

= 6 x 90°

= 540°

It is given that in a pentagon, two angles are 40° and 60°, and the remaining three angles are in the ratio 1 : 3 : 7.

Let the pentagon be ABCDE, such that ∠A = 40° and ∠B = 60°

Let the common factor of the remaining angles be a. Thus, the remaining angles are:

∠C = a, ∠D = 3a and ∠E = 7a.

Therefore,

⇒ ∠A + ∠B + ∠C + ∠D + ∠E = 540°

⇒ 40° + 60° + a + 3a + 7a = 540°

⇒ 100° + 11a = 540°

⇒ 11a = 540° - 100°

⇒ 11a = 440°

⇒ a = $\dfrac{440°}{11}$

⇒ a = 40°

Thus:

∠C = a = 40°

∠D = 3a = 3 x 40° = 120°

∠D = 7a = 7 x 40° = 280°

Hence, the largest angle is 280°.