Apply triangle angle sum and interior/exterior angle sums of polygons, including the angle properties of parallelograms and other special quadrilaterals.
Free, timed SHSAT practice sets on angles in triangles & polygons— every question includes a full worked explanation, and there's no account required.
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Start from two facts and you can answer almost any polygon angle question.
To find the number of sides from an interior angle, use the exterior angle: interior \(140^\circ\) means exterior \(40^\circ\), so \(n = 360 \div 40 = 9\).
Common traps: using \(n \times 180\) (forgetting the \(-2\)), giving the total when the question asks for each angle, and dividing 180 instead of 360 for exterior angles.
Interior angle sum \(= (n - 2)180 = (8 - 2)(180) = 1080^\circ\). (Forgetting the \(-2\), \(8 \times 180\), gives \(1440^\circ\). Using the exterior-angle sum gives \(360^\circ\). Dividing the sum by 8 gives one angle of a regular polygon, \(135^\circ\).)
Each interior angle \(= \dfrac{(n-2)180}{n} = \dfrac{(6-2)(180)}{6} = 120^\circ\). (Stopping at the sum, \((6-2)(180)\), and not dividing by 6 gives \(720^\circ\). Finding the exterior angle, \(360 \div 6\), gives \(60^\circ\). Subtracting the exterior angle from 360 instead of 180 gives \(300^\circ\).)
Each exterior angle is \(180 - 150 = 30^\circ\), and the exterior angles add to \(360^\circ\), so \(n = 360 \div 30 = 12\). (Stopping at the exterior angle gives 30. Dividing 180 instead of 360 by 30 gives 6. Subtracting 2 at the end, as in \((n - 2)180\), gives 10.)