Why a triangle's angles add to 180°
Tonight's idea
Take any triangle \(ABC\) and draw the line through \(C\) parallel to \(AB\). That one line is the whole proof.
Parallel lines make equal angles with any line that crosses them. In the figure, \(CA\) crosses both parallels, so the angle it makes with the new line at \(C\) equals angle \(A\). These are alternate angles, the two angles inside a letter Z. The same goes for \(CB\) and angle \(B\).
Now look at \(C\). The three angles there, \(a\), \(c\) and \(b\), sit side by side along a straight line, and a straight line is 180°. So
\(a + b + c = 180°\) for every triangle. No measuring needed.
Two things follow at once. Extend side \(AC\) beyond \(C\). The exterior angle there is \(180° - c\), which is \(a + b\): an exterior angle equals the sum of the two interior angles opposite it.
And a polygon with \(n\) sides can be cut into \(n - 2\) triangles by drawing diagonals from one corner, so its angles add to \((n - 2) \times 180°\). In a regular polygon all angles are equal, so divide by \(n\).
Watch one: a triangle, then a 20-gon
A triangle has angles 50° and 60°. The third is \(180° - 50° - 60° = 70°\), and the exterior angle at that vertex is \(180° - 70° = 110°\), which is indeed \(50° + 60°\).
A regular 20-gon: diagonals from one corner cut it into 18 triangles, so the angles add to \(18 \times 180° = 3240°\). Each of the 20 equal angles is \(3240° / 20 = 162°\). Each exterior angle is \(180° - 162° = 18°\), and \(20 \times 18° = 360°\): one full turn, which is no accident.
Drag the crossing line and watch the equal angles: mathdetours.com/demos/angles.html.
Your turn
- A triangle has angles 37° and 85°. Find the third angle, and the exterior angle at the third vertex.
- (a) An isosceles triangle has its two equal angles at the base and a 40° angle at the top. Find the base angles. (b) In a right-angled triangle the two smaller angles differ by 20°. Find them.
- Find the angle sum of a hexagon and of an octagon, and the size of each angle in a regular hexagon and a regular octagon. Regular hexagons fit together to tile a floor with no gaps and regular octagons do not. Use your angles to explain why.
- (a) Each angle of a regular polygon is 156°. How many sides does it have? (Hint: the exterior angles of any polygon add to 360°.) (b) Can a regular polygon have angles of exactly 155°? Explain.
Stretch: walking round, and a star
Prove that the exterior angles of any \(n\)-sided polygon add to 360°: each interior and exterior pair makes 180°, so use the interior sum. Then draw a pentagram, a five-pointed star. Each point is a triangle sitting on one side of the inner pentagon. Find the sum of the five point angles, for any star drawn this way, not only a regular one.
Answers
- \(180° - 37° - 85° = 58°\). Exterior angle \(180° - 58° = 122° = 37° + 85°\).
- (a) \((180° - 40°)/2 = 70°\) each. (b) They add to 90° and differ by 20°, so \(55°\) and \(35°\).
- Hexagon \(4 \times 180° = 720°\), each angle 120°. Octagon \(6 \times 180° = 1080°\), each angle 135°. Around a point there is 360°: three hexagon corners make exactly \(3 \times 120° = 360°\), but \(360° / 135°\) is not a whole number, so octagon corners cannot fill the space around a point. Two leave a 90° gap, three overlap.
- (a) Exterior angle \(180° - 156° = 24°\), and \(360° / 24° = 15\) sides. (b) The exterior angle would be 25°, and \(360 / 25 = 14.4\), not a whole number. So no.
Stretch. The \(n\) interior and exterior pairs total \(n \times 180°\). Take away the interior sum \((n - 2) \times 180°\), leaving \(2 \times 180° = 360°\). The pentagram: each point triangle has two base angles that are exterior angles of the inner pentagon, one at each end of its side. So the five point angles total \(5 \times 180° - (\text{all ten exterior angles})\). Going round the pentagon, each exterior angle appears twice, once for each neighbouring point, so the ten add to \(2 \times 360° = 720°\). Total \(900° - 720° = 180°\). In the regular case each point is 36°.