Smile 發問於 科學及數學數學 · 2 月前

Maths problem: how to do (a), thanks?

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  • 2 月前
    最愛解答

    (a)

    Let n be the number of sides of the polygon.

    Then angle sum of interior angles is (n - 2) × 180°.

    It is given that the largest interior angle is T(n) = 132°.

    Therefore, the smallest interior angle is T(1) = T(n) - (n-1)(12°) = 132° - (n-1)(12°).

    Equating the total sum of n interior angles is

    [T(1) + T(n)] × n / 2 = (n - 2) × 180°

    [264° - (n-1)(12°)] × n / 2 = (n - 2) × 180°

    (264 - 12n + 12)(n) = (n - 2)(360)

    264n - 12n² + 12n = 360n - 720

    12n² + 84n - 720 = 0

    n² + 7n - 60 = 0

    (n - 5)(n + 12) = 0

    n = 5  or  n = -12 (rejected).

    That is, the polygon is a pentagon with 5 sides.

    (b)

    The smallest interior angle is 132° - 4(12°) = 84°.

    Therefore, the second smallest interior angle is 84° + 12° = 96°.

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