QuanticaPrealgebraOpen in the courseYou can measure a single angle, and you know how angles pair up when parallel lines are involved. Now look at the angles inside a closed shape. They are not independent of each other. Fix all but one of them and the last one is forced. The triangle shows this most clearly. Its three angles always total the same amount no matter how you stretch or tilt it, and once you see why, the same reasoning carries up to a shape with any number of sides.
Problem
In the diagram above, a triangle has two angles marked \(47^\circ\) and \(72^\circ\). Find the third angle. Give the number of degrees.
Show a hint
- The three angles of a triangle always add to \(180^\circ\). Add the two you know, then see how much is left.
- Subtract both known angles from \(180^\circ\). Start with \(180 - 47\), then take away the \(72\).
Show the full solution
Two of the three angles are \(47^\circ\) and \(72^\circ\), so the third is whatever is left of \(180^\circ\). $$180 - 47 - 72 = \boxed{61}$$ Every triangle's angles add to \(180^\circ\), so knowing two of them always pins down the third.
Problem
In the diagram above, an isosceles triangle has its apex angle marked \(40^\circ\). The tick marks show the two sides meeting at the apex are equal, so the two base angles are equal. Find one base angle. Give the number of degrees.
Show a hint
- The three angles of any triangle add to \(180^\circ\). Subtract the apex angle to find how much the two base angles share between them.
- The tick marks tell you the base angles are equal, so split the leftover evenly between the two of them.
Show the full solution
The three angles add to \(180^\circ\), so the two base angles share \(180 - 40 = 140\) degrees, and the tick marks make them equal. $$\frac{180^\circ - 40^\circ}{2} = \boxed{70^\circ}$$ Equal sides force equal opposite angles, which is what lets you halve the leftover instead of guessing at a split.
Problem
In the diagram above, one side of a triangle is extended, forming an exterior angle of \(110^\circ\). One of the two remote interior angles is \(65^\circ\). Find the other remote interior angle. Give the number of degrees.
Show a hint
- The exterior angle theorem says an exterior angle equals the sum of the two remote interior angles, the two corners that do not touch it.
- Set \(65^\circ\) plus the unknown angle equal to \(110^\circ\), then solve for the unknown.
Show the full solution
The exterior angle equals the sum of the two remote interior angles, so the missing one is $$110^\circ - 65^\circ = \boxed{45}^\circ.$$ That is the exterior angle theorem. It works because the exterior angle and the third interior angle sit on a straight line.
Problem
In the diagram above, the three angles of a triangle are marked \(x^\circ\), \(2x^\circ\), and \(3x^\circ\). Find the largest angle. Give the number of degrees.
Show a hint
- The three angles of a triangle add to \(180^\circ\). Add the three marked pieces and set the total equal to \(180\).
- Once you solve for \(x\), the largest angle is the one marked \(3x^\circ\), so multiply your \(x\) by \(3\).
Show the full solution
The three angles of a triangle add to \(180^\circ\), so the marked pieces satisfy $$x + 2x + 3x = 180.$$ That gives \(6x = 180\), so \(x = 30\). The largest angle is the one marked \(3x^\circ\), which is \(3 \times 30 = \boxed{90}\).
Problem
In the diagram above is a hexagon. Find the sum of its interior angles. Give the number of degrees.
Show a hint
- The interior angles of any polygon add to \((n-2)\times 180^\circ\). Count the sides of the hexagon above to find \(n\).
- A hexagon has 6 sides, so \(n=6\). Work out \((6-2)\times 180^\circ\).
Show the full solution
A hexagon has 6 sides, so \(n=6\) in the interior angle sum \((n-2)\times 180^\circ\). $$(6-2)\times 180^\circ = 4\times 180^\circ = \boxed{720^\circ}$$ The \(n-2\) counts the triangles you get by drawing every diagonal from one vertex, and each triangle carries \(180^\circ\).
Problem
In the diagram above, a pentagon has four interior angles marked \(100^\circ\), \(110^\circ\), \(120^\circ\), and \(130^\circ\). Find the fifth angle. Give the number of degrees.
Show a hint
- A pentagon has 5 sides, so its interior angles add to \((5-2)\times 180^\circ\). Work out that total first.
- Subtract the four marked angles from that total to find the missing fifth angle.
Show the full solution
A pentagon's interior angles add to \((5-2)\times 180^\circ = 540^\circ\). Subtract the four marked angles from that total. $$540^\circ - 100^\circ - 110^\circ - 120^\circ - 130^\circ = \boxed{80}^\circ$$ Any polygon with one angle missing works this way, since the total is fixed by the side count alone.
Problem
The interior angles of a polygon add to \(1080^\circ\). How many sides does it have? Give the number of sides.
Show a hint
- The interior angles of an \(n\)-sided polygon add to \((n-2)\times 180^\circ\). Set that equal to \(1080^\circ\).
- Divide both sides by \(180\) to find \(n-2\), then add \(2\) to get \(n\).
Show the full solution
The interior angles of an \(n\)-sided polygon add to \((n-2)\times 180^\circ\), so set that total equal to \(1080^\circ\). $$(n-2)\times 180 = 1080$$ Divide both sides by \(180\) to get \(n-2 = 6\), then add \(2\) to both sides. The polygon has \(\boxed{8}\) sides.
Problem
The diagram above shows a regular octagon with one interior angle marked. Find that angle. Give the number of degrees.
Show a hint
- An octagon has \(8\) sides, so use the interior-angle formula for a regular polygon, \(\frac{(n-2)\times 180^\circ}{n}\), with \(n=8\).
- First find \((8-2)\times 180^\circ\), the total of all eight angles, then split it evenly among the \(8\) corners.
Show the full solution
A regular polygon has equal angles, so each interior angle is \(\frac{(n-2)\times 180^\circ}{n}\). With \(n=8\), the eight angles together add to \((8-2)\times 180^\circ = 1080^\circ\). Sharing that evenly among the \(8\) corners gives $$\frac{1080^\circ}{8} = \boxed{135}^\circ.$$
Problem
The diagram above shows a regular 20-sided polygon with one interior angle marked. Find that angle. Give the number of degrees.
Show a hint
- A regular polygon has all its interior angles equal, so each one is the total interior angle sum shared out evenly. That total is \((n-2)\times 180^\circ\), and here \(n=20\). Work out the total first.
- Once you have the sum \((20-2)\times 180^\circ = 3240^\circ\) for all twenty corners, one corner is just an equal slice of it. Divide by how many corners there are.
Show the full solution
A regular 20-gon splits its interior angle total evenly across all 20 corners. $$\frac{(20-2)\times 180^\circ}{20} = \frac{3240^\circ}{20} = \boxed{162}^\circ$$ The formula \(\frac{(n-2)\times 180^\circ}{n}\) is just the polygon angle sum divided by the number of equal corners.
Problem
In the diagram above, a regular pentagon has one exterior angle marked. Find that exterior angle. Give the number of degrees.
Show a hint
- The exterior angles of any polygon always add to \(360^\circ\), no matter how many sides.
- A regular pentagon has 5 equal corners, so split that total evenly.
Show the full solution
The exterior angles of any polygon add to \(360^\circ\), and a regular pentagon's 5 exterior angles are equal. $$\frac{360^\circ}{5} = \boxed{72}^\circ$$ That \(360^\circ\) never changes with the side count, so more sides just means smaller exterior angles.
Problem
In the diagram above, one corner of a regular polygon is shown, and each interior angle is \(150^\circ\). How many sides does the polygon have? Give the number of sides.
Show a hint
- At each corner the interior angle and the exterior angle sit on a straight line, so they add to \(180^\circ\). Find the exterior angle first.
- Walking once around a polygon, the exterior angles add up to \(360^\circ\). Divide that total by one exterior angle to count the corners.
Show the full solution
The interior and exterior angle at a corner form a straight line, so the exterior angle is \(180^\circ - 150^\circ = 30^\circ\). The exterior angles of any polygon add to \(360^\circ\), and here they are all equal, so$$n = \frac{360^\circ}{30^\circ} = \boxed{12}.$$
Problem
In the diagram above, a regular pentagon and a regular hexagon share a side and meet at a point, leaving a gap. Find the gap angle marked at that point. Give the number of degrees.
Show a hint
- The three angles around the shared point make a full turn, so they add to \(360^\circ\). Two of them are the corner angles of the polygons.
- A regular pentagon has interior angle \(\frac{(5-2)\times 180^\circ}{5}\) and a regular hexagon \(\frac{(6-2)\times 180^\circ}{6}\). Subtract both from \(360^\circ\).
Show the full solution
The pentagon's corner is \(\frac{(5-2)\times 180^\circ}{5}=108^\circ\) and the hexagon's is \(\frac{(6-2)\times 180^\circ}{6}=120^\circ\). The three angles around the shared point make a full turn, so the gap is what is left. $$360^\circ - 108^\circ - 120^\circ = \boxed{132}^\circ$$ Since \(108 + 120\) falls short of \(360\), a pentagon and a hexagon can never tile a flat surface together around a point.
Problem
In the diagram above, one corner of a regular polygon is shown where each interior angle is \(4\) times its exterior angle. How many sides does the polygon have? Give the number of sides.
Show a hint
- At any corner the interior and exterior angle sit on a straight line, so they add to \(180^\circ\). Call the exterior angle \(e\) and write the interior angle as \(4e\).
- Once you know one exterior angle, the number of sides comes from \(\frac{360^\circ}{n} = e\).
Show the full solution
The interior and exterior angle at a corner form a linear pair, so they add to \(180^\circ\). The interior angle is \(4\) times the exterior angle, so with exterior angle \(e\), $$4e + e = 180^\circ,\qquad 5e = 180^\circ,\qquad e = 36^\circ.$$ Every exterior angle of a regular polygon is \(\frac{360^\circ}{n}\), so \(\frac{360^\circ}{n} = 36^\circ\), which gives \(n = \boxed{10}\).
Practice these ideas
Practice
In the diagram above, a triangle has angles \(55^\circ\) and \(80^\circ\). Find the third angle. Give the number of degrees.
Show the solution
Two angles are \(55^\circ\) and \(80^\circ\), so the third is what is left of \(180^\circ\). $$180^\circ - 55^\circ - 80^\circ = \boxed{45}$$
Practice
In the diagram above, an isosceles triangle has an apex angle of \(100^\circ\), and the tick marks show that the two base angles are equal. Find one base angle. Give the number of degrees.
Show the solution
The angles in a triangle add to \(180^\circ\), so the two base angles together make \(180^\circ - 100^\circ = 80^\circ\). The tick marks tell us those base angles are equal, so each one is half of that. $$\frac{80^\circ}{2} = \boxed{40}$$
Practice
In the diagram above, one side of a triangle is extended to make an exterior angle of \(125^\circ\). One of the two remote interior angles is \(60^\circ\). Find the other remote interior angle. Give the number of degrees.
Show the solution
By the exterior angle theorem, an exterior angle equals the sum of the two remote interior angles. So the two remote angles add to \(125^\circ\). One of them is \(60^\circ\), so the other is $$125^\circ - 60^\circ = \boxed{65^\circ}.$$
Practice
In the diagram above, a triangle has angles \(2x^\circ\), \(3x^\circ\), and \(4x^\circ\). Find the largest angle. Give the number of degrees.
Show the solution
The angles in a triangle sum to \(180^\circ\), so add the three expressions and set them equal to \(180\). $$2x + 3x + 4x = 180$$ That gives \(9x = 180\), so \(x = 20\). The largest angle is \(4x\), which is \(4\times 20 = \boxed{80}\).
Practice
The diagram above shows a heptagon (a polygon with 7 sides). Find the sum of its interior angles. Give the number of degrees.
Show the solution
A heptagon has \(n=7\) sides, so use \((n-2)\times 180^\circ\). $$(7-2)\times 180^\circ = 5\times 180^\circ = \boxed{900}^\circ$$ Seven sides means five triangles fan out from one vertex, and each triangle carries \(180^\circ\).
Practice
In the diagram above, five of the six interior angles of a hexagon are \(110^\circ\), \(130^\circ\), \(115^\circ\), \(125^\circ\), and \(120^\circ\). Find the sixth angle. Give the number of degrees.
Show the solution
A hexagon has \(n=6\) sides, so its interior angles add to \((6-2)\times 180^\circ = 720^\circ\). The five marked angles add to $$110^\circ + 130^\circ + 115^\circ + 125^\circ + 120^\circ = 600^\circ.$$ The sixth angle is what is left over, so \(720^\circ - 600^\circ = \boxed{120}\) degrees.
Practice
The interior angles of a polygon add to \(1440^\circ\). How many sides does it have? Give the number of sides.
Show the solution
By the polygon angle sum, the interior angles of an \(n\)-sided polygon add to \((n-2)\times 180^\circ\). Set this equal to the total. $$(n-2)\times 180^\circ = 1440^\circ$$ Divide both sides by \(180^\circ\) to get \(n-2 = 8\), so \(n = \boxed{10}\).
Practice
The diagram above shows a regular dodecagon, a polygon with \(12\) equal sides and \(12\) equal angles. Find the measure of one interior angle. Give the number of degrees.
Show the solution
A regular polygon spreads its interior angle total evenly across all \(n\) corners, so each angle is \(\frac{(n-2)\times 180^\circ}{n}\). Here \(n = 12\). $$\frac{(12-2)\times 180^\circ}{12} = \frac{1800^\circ}{12} = \boxed{150^\circ}$$
Practice
In the diagram above, a regular polygon has 15 sides, with one interior angle marked. Find the size of that angle. Give the number of degrees.
Show the solution
The interior angles of any polygon add to \((n-2)\times 180^\circ\), and a regular polygon splits that total evenly across all \(n\) angles. With \(n=15\), $$\frac{(15-2)\times 180^\circ}{15} = \frac{2340^\circ}{15} = \boxed{156^\circ}.$$
Practice
In the diagram above, a regular decagon has one exterior angle marked. Find that exterior angle. Give the number of degrees.
Show the solution
A regular decagon has \(n=10\) equal exterior angles, and the exterior angles of any polygon add to \(360^\circ\). $$\frac{360^\circ}{10} = \boxed{36}$$ That \(360^\circ\) is the full turn you make walking once around the shape, so it holds for every polygon.
Practice
In the diagram above, one corner of a regular polygon has an exterior angle of \(45^\circ\), where a side is extended past the vertex. How many sides does the polygon have? Give the number of sides.
Show the solution
The exterior angles of a regular polygon are all equal and add to \(360^\circ\), so each one is \(\frac{360^\circ}{n}\). Here that angle is \(45^\circ\). $$\frac{360^\circ}{n}=45^\circ \;\Rightarrow\; n=\frac{360}{45}=\boxed{8}$$ Reading the rule backwards like this turns any exterior angle straight into a side count.
Practice
The diagram above shows one corner of a regular polygon, where each interior angle is \(140^\circ\). How many sides does the polygon have? Give the number of sides.
Show the solution
The interior and exterior angles sit on a straight line, so the exterior angle is \(180^\circ - 140^\circ = 40^\circ\). $$n = \frac{360^\circ}{40^\circ} = \boxed{9}$$ Going through the exterior angle is much faster than solving \(\frac{(n-2)\times 180}{n} = 140\) directly.
Practice
In the diagram above, one corner of a regular polygon is marked, and its interior angle is \(5\) times its exterior angle. How many sides does the polygon have? Give the number of sides.
Show the solution
Interior and exterior angles at a corner form a linear pair, so they add to \(180^\circ\). The interior angle is \(5\) times the exterior angle, so with exterior angle \(e\), $$5e + e = 180^\circ,\quad 6e = 180^\circ,\quad e = 30^\circ.$$ For a regular polygon the exterior angle is \(\frac{360^\circ}{n}\), so $$\frac{360^\circ}{n} = 30^\circ \;\Rightarrow\; n = \frac{360^\circ}{30^\circ} = \boxed{12}.$$