Four of the five interior angles of a pentagon are \(100^\circ\), \(120^\circ\), \(90^\circ\) and \(130^\circ\). Work out the size of the fifth angle.
Question 1 Report
Four of the five interior angles of a pentagon are \(100^\circ\), \(120^\circ\), \(90^\circ\) and \(130^\circ\). Work out the size of the fifth angle.
Answer Details
The sum of the interior angles of a pentagon is \((5-2) \times 180^{\circ} = 540^{\circ}\) [M1].
The four known angles add to \(100^{\circ}+120^{\circ}+90^{\circ}+130^{\circ} = 440^{\circ}\), so the fifth angle is \(540^{\circ} - 440^{\circ}\) [M1], giving \(100^{\circ}\) [A1].
Exam tip: always find the total sum for the polygon first using \((n-2)\times180^{\circ}\), then subtract the known angles - never assume a pentagon's angles sum to \(360^{\circ}\), which is only true for angles at a point or exterior angles.
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