The sum of the squares of two consecutive multiplies of 7 is 637.

Taking the bigger number x a positive number, find the smaller of these two angles.

Let the multiplies of 7 be 7x and 7(x + 1).

It is given that the sum of the squares of two consecutive multiplies of 7 is 637.

⇒ (7x)² + [7(x + 1)]² = 637

⇒ 49x² + [7x + 7]² = 637

⇒ 49x² + 49x² + 7² + 98x - 637 = 0

⇒ 98x² + 49 + 98x - 637 = 0

⇒ 98x² + 98x - 588 = 0

⇒ x² + x - 6 = 0

⇒ x² + 3x - 2x - 6 = 0

⇒ x(x + 3) - 2(x + 3) = 0

⇒ (x + 3)(x - 2) = 0

⇒ (x + 3) = 0 or (x - 2) = 0

⇒ x = -3 or x = 2

It is given that x is bigger positive number.

So, the multiplies of 7 is 7 x 2 = 14 and 7(2 + 1) = 7 x 3 = 21

Hence, the smaller angle is 14.