Question

Two interior angles $A$ and $B$ of pentagon $ABCDE$ are $60^{\circ}$ and $85^{\circ}$. Two of the remaining angles, $C$ and $D$, are equal and the fifth angle $E$ is $15^{\circ}$ more than twice $C$. Find the measure of the largest angle.

Answers

  1. The largest angle is E = 205°.

    What is a pentagon?

    • A pentagon is any five-sided polygon or 5-gon in geometry.
    • A basic pentagon’s internal angles add up to 540°.
    • A pentagon might be straightforward or self-intersecting.
    • A pentagram is a self-intersecting regular pentagon (or a star pentagon).
    To find the measure of the largest angle:
    The sum of interior angles of a polygon is:
    • Sum = 180° × (n – 2)
    Where n is the number of sides.
    Here n = 5, as pentagon has 5 sides.
    • Sum = 180° × 3 = 540°
    Let, C = D = x
    Then, E = 2x + 15 (15 more than twice C)
    We can express the sum of the 5 angles as:
    • 60 + 85 + x + x + 2x + 15 = 540
    Simplify the left side by collecting like terms:
    • 4x + 160 = 540
    Subtract 160 from both sides:
    • 4x + 160 – 160 = 540 – 160
    • 4x = 380
    Divide both sides by 4:
    • 4x/4 = 380/4 = x = 95
    • C = D = x = 95°
    • E (2 × 95) + 15 = 205°
    The 5 interior angles of the pentagon are:
    • 60° + 85° + 95° + 95° + 205° = 540°
    The largest angle is E = 205°.
    Therefore, the largest angle is E = 205°.
    Know more about a pentagon here:
    #SPJ4

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