Triangle angle sum
Triangles come in every shape, long and thin or short and wide. Is there anything about their angles, inside the corners or just outside them, that every one of them shares?
Where you are 18 in Triangles
Why this exists
A triangle has three sides and three angles, and it can be almost any shape. A tall thin triangle has one tiny angle at the top and two steep ones at the bottom. A flat wide triangle has one huge angle and two small ones. It looks as if the three angles could be anything at all.
They cannot. Whatever the shape and whatever the size, the three angles of a triangle always add to the same total, 180^\circ, the size of a straight angle. So two angles of a triangle always tell you the third, without a protractor. That one fact is behind a large share of the angle work in geometry, from isosceles triangles to the angles of any polygon.
Many of the angles you meet in a figure are not inside the triangle at all. Whenever a triangle sits on a longer line, one of its sides runs on past a corner, and a new angle opens up outside. That outside angle looks like new information. It is not. It is fixed by the two corners at the far end of the triangle, by a rule you can use at a glance: add them. The rule skips a step in every problem, and, more usefully, it lets an angle belonging to one triangle be read straight off another triangle that shares a side with it. That is how long angle chases become short ones.
Definition
Interior angles and the angle sum
The angles of a triangle are its three interior angles: the angle at each corner, between the two sides that meet there, on the inside. In \triangle ABC they are \angle A, \angle B and \angle C.
The angle sum of a triangle is the total of their three measures:
The theorem below says this total is the same for every triangle.
Exterior angles
Extend one side of a triangle beyond a vertex. The angle between the extension and the other side at that vertex is an exterior angle of the triangle.
In Figure 1, side \overline{BC} of \triangle ABC is extended past C to D, and \angle 4, which is \angle ACD, is an exterior angle at C.
The three angles of the triangle stand in two different relations to \angle 4.
- \angle 3, the inside angle at the same vertex, is the adjacent interior angle. It shares the side \overline{CA} with \angle 4, and the two fill the straight line through B, C and D, so they form a linear pair.
- \angle 1 and \angle 2, at the other two vertices, are the remote interior angles of \angle 4. Remote means far away: neither one touches C. Some books call them the non-adjacent or opposite interior angles.
Each vertex has two exterior angles, one for each side you could extend. They are vertical angles, so they are equal, and “the exterior angle at C” means the same number whichever one is drawn. The lesson on complementary and supplementary angles shows this from the other direction: both are supplements of \angle 3.
Intuition
Cut a triangle out of paper and tear off its three corners. Lay them side by side so that they share one point and do not overlap. However the triangle was drawn, the three corners fit together into a straight line.
Now picture a triangle changing shape. Keep the base \overline{BC} fixed and push the top vertex A down toward the base. The angle at A opens up toward a straight angle, and the angles at B and C shrink toward nothing. Pull A far up instead and the angle at A closes toward nothing while the two base angles open toward right angles. Whatever one angle gains, the other two lose. The total never moves.
Tearing paper shows that it happens for the triangles you tried. It does not show why, or that it must happen for every triangle. The reason comes under “Why it is true”.
The angle outside a corner
Euclid saw the exterior angle rule directly, without adding up the triangle first. Go back to Figure 1 and draw through C the ray \overrightarrow{CE} parallel to \overline{BA}, as in Figure 3. It cuts the exterior angle into two pieces.
The piece next to \overrightarrow{CD} matches \angle 2: the line through B and D crosses the two parallels, and the two angles are corresponding angles. The piece next to \overline{CA} matches \angle 1: the segment \overline{AC} crosses the same two parallels, and those two are alternate interior angles.
So the outside angle is made of the two far corners, slid over to C and set side by side. Move A anywhere above the line and all of this moves together: whatever \angle 1 and \angle 2 become, the exterior angle is their total. This is Euclid’s own construction in Proposition 32, a route to the exterior angle theorem that never passes through the angle sum. It is also the torn paper again: the two copies at C and \angle 3 beside them fill the straight line through B, C and D.
Results
Triangle Angle Sum Theorem. The three angles of every triangle add to 180^\circ.
Several facts follow at once.
- The third angle. Two angles of a triangle fix the third: m\angle C = 180^\circ - m\angle A - m\angle B.
- Right triangles. The right angle uses 90^\circ, so the other two angles add to 90^\circ: they are complementary.
- At most one big angle. A triangle has at most one right angle or obtuse angle, so it has at least two acute angles. This is why classifying a triangle by its angles only has to look at its largest one.
- Equilateral triangles. Three equal angles each measure 180^\circ \div 3 = 60^\circ.
- Two angles match, so the third matches. If two angles of one triangle equal two angles of another, the third angles are equal too.
Exterior Angle Theorem. An exterior angle of a triangle equals the sum of its two remote interior angles. In Figure 1,
Exterior Angle Inequality. An exterior angle of a triangle is greater than each of its remote interior angles:
Some books, following Euclid, give the name “exterior angle theorem” to this inequality instead.
The angle beside it. An exterior angle and its adjacent interior angle are supplementary:
Why it is true
The angle sum
The reason uses parallel lines cut by a transversal, and one fact about them in particular, alternate interior angles: when a transversal cuts two parallel lines, the angles tucked into the two bends of the Z are equal.
Take any triangle ABC. Through the vertex A, draw the line l parallel to the opposite side \overline{BC}. Such a line can always be drawn, and by the parallel postulate there is only one. That “only one” is what the next step leans on: the fact that alternate interior angles at parallel lines are equal is itself proved from the parallel postulate, so this theorem depends on the postulate too.
At A there are now three angles side by side along l: \angle 1 on the left, the triangle’s own angle \angle 2 in the middle, and \angle 3 on the right. (The numbers in Figure 4 name its own angles, not the ones in Figure 1.) Together they fill the straight angle on one side of l, so
The side \overline{AB} is a transversal of the parallel lines l and \overleftrightarrow{BC}. The angles \angle 1 and \angle 4 both sit between those lines, on opposite sides of \overline{AB}, so they are alternate interior angles, and m\angle 1 = m\angle 4.
In the same way, \overline{AC} is a transversal, and \angle 3 and \angle 5 are alternate interior angles, so m\angle 3 = m\angle 5.
Replace \angle 1 and \angle 3 in the straight-angle equation by the angles they equal:
These are exactly the three angles of the triangle, at B, A and C.
Nothing in the argument depended on the triangle’s shape. It works for acute, right and obtuse triangles alike, which is what the torn paper could not show. It is also the reason the corners fit in Figure 2: the parallel line lays the two base angles beside the top one. The result is Proposition 32 of Book I of Euclid’s Elements.
The exterior angle theorem
Two facts already known, and one subtraction. The angle numbers here are the ones in Figure 1.
\angle 3 and \angle 4 form a linear pair, so
\angle 1, \angle 2 and \angle 3 are the three angles of \triangle ABC, so by the triangle angle sum
Both left sides equal 180^\circ, so they equal each other:
Subtract m\angle 3 from both sides:
The inequality follows at once. \angle 2 is a real angle, so m\angle 2 > 0, and therefore m\angle 4 = m\angle 1 + m\angle 2 is more than m\angle 1. Swap the roles of \angle 1 and \angle 2 and the same sentence shows m\angle 4 > m\angle 2.
The move is the one that proves vertical angles equal: \angle 4 and the pair \angle 1, \angle 2 are both whatever \angle 3 leaves over from 180^\circ, so they match. This runs Euclid’s order backwards: he proves the exterior angle theorem from the parallel in Figure 3 and then gets the angle sum from it, both in Proposition 32 of Book I of the Elements.
Written out as formal two-column proofs, both theorems are their own skill: proving the triangle angle sum, which also proves the exterior angle theorem.
Worked examples
Two angles given
Find x in Figure 5.
The three angles of the triangle add to 180^\circ, so
The two known angles add to 119, so x = 180 - 119 = 61. The third angle measures 61^\circ.
Check. 48 + 71 + 61 = 180.
An acute angle of a right triangle
One acute angle of a right triangle measures 34^\circ. Find the other acute angle.
SolutionThe right angle takes 90^\circ of the 180^\circ, so the two acute angles share the remaining 90^\circ:
The other acute angle measures 56^\circ. The two acute angles of every right triangle are complementary for this reason.
All three angles in terms of x
The angles of the triangle in Figure 6 are written as expressions in x. Find x and all three angles.
The three expressions are the three angles of one triangle, so they add to 180:
Substitute to get the angles: 32 + 15 = 47, 2(32) = 64 and 3(32) - 27 = 69. The angles are 47^\circ, 64^\circ and 69^\circ.
Check. 47 + 64 + 69 = 180, and every angle is positive, so this is a real triangle.
Angles in a ratio
The angles of a triangle are in the ratio 2 : 3 : 7. Find them, and say whether the triangle is acute, right or obtuse.
SolutionA ratio of 2 : 3 : 7 means the angles are 2k, 3k and 7k degrees for some number k. They add to 180:
So the angles are 30^\circ, 45^\circ and 105^\circ. The largest is more than 90^\circ, so the triangle is obtuse.
An isosceles triangle, two ways
One angle of an isosceles triangle measures 40^\circ. Find the other two angles.
SolutionAn isosceles triangle has two equal angles, the base angles, opposite its two equal sides (the base angles theorem). The third angle is the vertex angle. The problem does not say which kind the 40^\circ angle is, so try both.
If 40^\circ is the vertex angle, the two equal base angles share the rest:
The angles are 40^\circ, 70^\circ and 70^\circ.
If 40^\circ is a base angle, the other base angle is also 40^\circ, and the vertex angle is
The angles are 40^\circ, 40^\circ and 100^\circ. Both triangles exist, so both answers stand.
A triangle cut in two
In Figure 7, D is a point on \overline{BC} and the segment \overline{AD} is drawn. Find x, y and z.
The segment makes two smaller triangles, \triangle ABD and \triangle ADC, and the theorem holds in each one.
In \triangle ABD the angles are 50^\circ at B, 30^\circ at A and x at D, so
The angles x and y sit side by side along the straight line \overline{BC}, so they form a linear pair:
In \triangle ADC the angles are z at A, 80^\circ at D and 60^\circ at C, so
Check. The whole angle at A is 30^\circ + 40^\circ = 70^\circ. In the big triangle \triangle ABC that angle must be 180^\circ - 50^\circ - 60^\circ = 70^\circ. The two routes agree.
Finding an exterior angle
In Figure 8, find x.
The angle marked x^\circ is the exterior angle at C. Its remote interior angles are the ones at A and B, so it equals their sum:
Check. Go the long way round. By the angle sum, the interior angle at C is 180^\circ - 48^\circ - 67^\circ = 65^\circ, and the exterior angle forms a linear pair with it: 180^\circ - 65^\circ = 115^\circ. The two routes agree. The exterior angle theorem is the route that skips the middle angle.
Finding a remote interior angle
An exterior angle of a triangle measures 128^\circ, and one of its remote interior angles measures 57^\circ. Find the other remote interior angle and the interior angle adjacent to the exterior angle.
SolutionCall the unknown remote angle r^\circ. The two remote angles add to the exterior angle:
The adjacent interior angle forms a linear pair with the exterior angle, so it is 180^\circ - 128^\circ = 52^\circ.
Check. The triangle’s three angles are 57^\circ, 71^\circ and 52^\circ, and 57 + 71 + 52 = 180.
An exterior angle equation in x
In Figure 9, find x and all three angles of the triangle.
\angle ACD is exterior at C, and its remote interior angles are \angle A and \angle B. The exterior angle theorem gives an equation:
Substitute back: m\angle A = 4(20) - 5 = 75^\circ, m\angle B = 3(20) + 10 = 70^\circ, and the exterior angle is 8(20) - 15 = 145^\circ. The interior angle at C is its supplement, 180^\circ - 145^\circ = 35^\circ.
Check. 75 + 70 = 145, and the triangle’s angles add to 75 + 70 + 35 = 180.
The exterior angle at the apex of an isosceles triangle
The exterior angle at the apex of an isosceles triangle measures 110^\circ. Find all three angles of the triangle.
SolutionThe two remote interior angles of the apex’s exterior angle are the two base angles, and by the base angles theorem they are equal. Call each b^\circ:
The apex angle is the supplement of the exterior angle, 180^\circ - 110^\circ = 70^\circ.
Check. 55 + 55 + 70 = 180. In any isosceles triangle the exterior angle at the apex is exactly twice a base angle.
An angle in one triangle read from another
In Figure 10, D lies on \overline{BC}. Find x and y.
Look for a triangle in which \angle ADC is exterior. In \triangle ABD, side \overline{BD} is extended past D to C, so \angle ADC is exterior to \triangle ABD at D. Its remote interior angles are \angle B and \angle BAD:
Now \triangle ADC has angles x^\circ = 70^\circ at D and 65^\circ at C, so by the angle sum
Check. The whole angle at A is 30^\circ + 45^\circ = 75^\circ, and in the big triangle 40 + 75 + 65 = 180. Example 6 reached the same kind of angle at D in two steps, the angle sum in \triangle ABD and then a linear pair; the exterior angle theorem does it in one.
One exterior angle at each vertex
In Figure 11, each side of the triangle is extended once, going around it. Show that x + y + z = 360, whatever the triangle.
Write a, b and c for the measures of the interior angles at A, B and C. Each exterior angle is the sum of the two interior angles not at its own vertex:
Add the three equations. Each interior angle appears twice:
The shape of the triangle never entered, so the total is always 360^\circ. The triangle in Figure 11 is drawn with a = 60, b = 65 and c = 55, which gives x = 115, y = 125 and z = 120, and 115 + 125 + 120 = 360.
Boundaries
Interior angles only in the sum. The 180^\circ total is about the three angles inside the triangle. An exterior angle is not one of them; it obeys its own rule, the exterior angle theorem.
An exterior angle needs an extended side. The angle outside a corner measured the long way round, 360^\circ minus the interior angle, is a reflex angle, not an exterior angle. An exterior angle is always the supplement of the angle beside it, so it lies strictly between 0^\circ and 180^\circ.
Remote means the other two. The interior angle at the same vertex is never part of the exterior angle’s sum. It is the supplement of the exterior angle, not a piece of it.
Triangles only. The exterior angle proof works because the linear pair and the angle sum both come to 180^\circ. In a quadrilateral the angles add to 360^\circ, and the three far angles add to more than the exterior angle (Exercise 45).
The sides must be straight and the surface flat. The proof of the angle sum uses alternate interior angles at parallel lines, and their equality rests on the parallel postulate. Where that postulate fails, so does the theorem. On the surface of a sphere, where the straightest paths are great circles, a triangle’s angles add to more than 180^\circ. Take the north pole and two points on the equator a quarter of the way round from each other: each of the three angles is a right angle, and the sum is 270^\circ. The exterior angle theorem rests on the angle sum, so it fails there too: an exterior angle comes out smaller than the sum of its two remote angles. The inequality asks for less. Euclid proves it as Proposition 16 of Book I, before he uses parallel lines at all, so it survives in geometries where the parallel postulate fails. It does not survive on a sphere, where lines close up into circles; a large enough spherical triangle has an exterior angle smaller than a remote one.
Angles fix the shape, not the size. A small triangle and a large one can have exactly the same three angles. Knowing the angles tells you nothing about how long the sides are. That is the start of similar triangles.
Every angle must be positive. Any three positive measures that add to 180^\circ are the angles of some triangle. Three numbers that add to 180 with a zero or a negative among them are not: 90^\circ, 90^\circ, 0^\circ describes two parallel sides that never meet.
Where this goes next
The theorem is usually one link in a longer chain. Combined with vertical angles, linear pairs and parallel lines, it lets you find every angle in a figure. The exterior angle theorem is the shortcut in nearly every such angle hunt with more than one triangle in it, as in Example 11.
Cut a quadrilateral along a diagonal and you get two triangles, so its angles add to 360^\circ (the angle sum of a quadrilateral); cut any polygon into triangles the same way and you get the interior angle sum of a polygon. Example 12 is the first case of a matching fact about the outside: walk once around any convex polygon and the exterior angles add to 360^\circ, which is the exterior angle sum of a polygon.
Because two angles fix the third, two triangles with two matching angles have all three matching, which is why angle angle similarity needs only two. The same fact gives the equal angles of the equilateral triangle and the 45^\circ and 60^\circ angles of the special right triangles.
The exterior angle inequality does quiet work in the theory of triangles, where it helps show that the larger angle faces the longer side. And in circles, most of the angle theorems are proved by spotting a triangle and one of its exterior angles: the inscribed angle theorem and the angles made by two chords, and by secants and tangents outside a circle all start that way.
Exercises
Finding the third angle
Two angles of a triangle are given. Find the third.
- 35^\circ and 85^\circ
- 112^\circ and 29^\circ
- 90^\circ and 47^\circ
- 60^\circ and 60^\circ
- 17.5^\circ and 101.5^\circ
Finding an exterior angle
The two remote interior angles of an exterior angle are given. Find the exterior angle.
- 35^\circ and 80^\circ
- 90^\circ and 27^\circ
- 62.5^\circ and 41^\circ
- Find an exterior angle of an equilateral triangle.
Working backwards from an exterior angle
- An exterior angle measures 140^\circ and one of its remote interior angles measures 85^\circ. Find the other remote interior angle.
- An exterior angle measures 96^\circ and its two remote interior angles are equal. Find them.
- An exterior angle measures 150^\circ, and its two remote interior angles are in the ratio 2 : 3. Find all three angles of the triangle.
- In \triangle ABC, side \overline{BC} is extended past C to D, m\angle ACD = 104^\circ and m\angle A = 38^\circ. Find m\angle B and m\angle ACB.
Using algebra
- The angles of a triangle are x^\circ, (3x)^\circ and (5x)^\circ. Find x and the three angles.
- The angles of a triangle are (x + 10)^\circ, (x + 20)^\circ and (x + 30)^\circ. Find x and the three angles.
- The angles of a triangle are (5x - 4)^\circ, (3x + 2)^\circ and (2x + 12)^\circ. Find x and the three angles.
- The two acute angles of a right triangle are (3y + 6)^\circ and (2y - 1)^\circ. Find y and both angles.
- The angles of a triangle are in the ratio 1 : 4 : 5. Find them, and say whether the triangle is acute, right or obtuse.
- One angle of a triangle is twice the smallest angle, and the third is 30^\circ more than the smallest. Find all three angles.
- The remote interior angles of an exterior angle measure (x + 15)^\circ and (2x - 10)^\circ, and the exterior angle measures (4x - 25)^\circ. Find x and all three measures.
- The remote interior angles measure (5x)^\circ and (3x + 6)^\circ, and the exterior angle measures (10x - 18)^\circ. Find x and all three measures.
- An exterior angle measures (7y - 12)^\circ and its remote interior angles measure (2y + 8)^\circ and (3y + 4)^\circ. Find y and all three measures.
- An exterior angle measures (x^2)^\circ and its remote interior angles measure (4x)^\circ and (3x + 18)^\circ. Find every value of x that gives genuine angles, and the measures.
- In a right triangle, the exterior angle at one acute vertex measures (3x + 15)^\circ and the other acute angle measures (x + 5)^\circ. Find x and the three angles of the triangle.
Isosceles triangles
- The vertex angle of an isosceles triangle is 96^\circ. Find each base angle.
- A base angle of an isosceles triangle is 38^\circ. Find the vertex angle.
- One angle of an isosceles triangle is 70^\circ. Find every possible pair of values for the other two angles.
- An isosceles triangle has a right angle. Find its other two angles.
In figures
- In a figure like Figure 7, m\angle B = 42^\circ, m\angle BAD = 58^\circ and m\angle C = 35^\circ. Find m\angle ADB, m\angle ADC and m\angle DAC.
- Three angles of a quadrilateral are 85^\circ, 110^\circ and 70^\circ. Split it into two triangles with a diagonal, and use the split to find the fourth angle.
- Segments \overline{AC} and \overline{BD} cross at E, making triangles ABE and CDE. m\angle BAE = 50^\circ, m\angle ABE = 65^\circ and m\angle ECD = 45^\circ. Find m\angle AEB and m\angle EDC.
- D lies on side \overline{BC} of \triangle ABC, and \overline{AD} is drawn. m\angle B = 35^\circ, m\angle BAD = 25^\circ and m\angle C = 70^\circ. Use the exterior angle theorem to find m\angle ADC, then find m\angle DAC.
- In \triangle PQR, PQ = PR and m\angle P = 40^\circ. Side \overline{QR} is extended past R to S. Find m\angle PRS.
- In \triangle ABC, AB = AC. Side \overline{BC} is extended past C to D, and m\angle ACD = 115^\circ. Find all three angles of the triangle.
- A triangle has one exterior angle drawn at each vertex, as in Figure 11. Two of them measure 110^\circ and 130^\circ. Find the third, and the three angles of the triangle.
Reasoning
- Can a triangle have two right angles? Explain.
- Explain why the two acute angles of any right triangle add to 90^\circ.
- In \triangle ABC and \triangle DEF, m\angle A = m\angle D and m\angle B = m\angle E. Show that m\angle C = m\angle F.
- Explain why every triangle has at least two acute angles.
- Someone measures the angles of a paper triangle with a protractor and gets 57^\circ, 61^\circ and 64^\circ. Does this show the theorem is wrong? Explain.
- Can an exterior angle of a triangle be acute? Can a triangle have acute exterior angles at two different vertices? Explain.
- Explain why an exterior angle of a triangle is always larger than each of its remote interior angles.
- An exterior angle of a triangle is a right angle. Show that the triangle is a right triangle and that its other two angles are complementary.
- In \triangle ABC, AB = AC. Side \overline{BA} is extended past A to E, and ray \overrightarrow{AF} bisects the exterior angle \angle CAE. Show that \overrightarrow{AF} is parallel to \overline{BC}.
- In quadrilateral ABCD, side \overline{CD} is extended past D. Is the exterior angle at D equal to m\angle A + m\angle B + m\angle C? Find the correct relation and explain it.
Answers
- 60^\circ. 180 - 35 - 85 = 60.
- 39^\circ. 180 - 112 - 29 = 39.
- 43^\circ. 180 - 90 - 47 = 43; the two acute angles of a right triangle add to 90^\circ.
- 60^\circ. 180 - 60 - 60 = 60, so all three angles are 60^\circ.
- 61^\circ. 17.5 + 101.5 = 119, and 180 - 119 = 61.
- 115^\circ, since 35^\circ + 80^\circ = 115^\circ.
- 117^\circ, since 90^\circ + 27^\circ = 117^\circ.
- 103.5^\circ, since 62.5^\circ + 41^\circ = 103.5^\circ.
- 120^\circ. Each angle of an equilateral triangle is 60^\circ, and an exterior angle is the sum of two of them: 60^\circ + 60^\circ = 120^\circ.
- 55^\circ. The remote angles add to the exterior angle, so 85 + r = 140.
- 48^\circ each. With each r^\circ, 2r = 96.
- 60^\circ, 90^\circ and 30^\circ. Write the remote angles as 2k and 3k; then 5k = 150 and k = 30. The third angle is beside the exterior angle, so it is 180^\circ - 150^\circ = 30^\circ.
- m\angle B = 66^\circ and m\angle ACB = 76^\circ. \angle A and \angle B are the remote angles of \angle ACD, so 38 + m\angle B = 104. \angle ACB forms a linear pair with \angle ACD, so it is 180^\circ - 104^\circ. Check: 38 + 66 + 76 = 180.
- x = 20; the angles are 20^\circ, 60^\circ and 100^\circ. From x + 3x + 5x = 180, 9x = 180.
- x = 40; the angles are 50^\circ, 60^\circ and 70^\circ. From (x + 10) + (x + 20) + (x + 30) = 180, 3x + 60 = 180 and 3x = 120.
- x = 17; the angles are 81^\circ, 53^\circ and 46^\circ. The sum is 10x + 10 = 180, so 10x = 170. Check: 81 + 53 + 46 = 180.
- y = 17; the angles are 57^\circ and 33^\circ. The acute angles add to 90^\circ: (3y + 6) + (2y - 1) = 90, so 5y + 5 = 90 and 5y = 85.
- 18^\circ, 72^\circ and 90^\circ, a right triangle. The angles are k, 4k and 5k, so 10k = 180 and k = 18.
- 37.5^\circ, 75^\circ and 67.5^\circ. With smallest angle x, x + 2x + (x + 30) = 180, so 4x = 150 and x = 37.5.
- x = 30: the remote angles are 45^\circ and 50^\circ and the exterior angle is 95^\circ. From (x + 15) + (2x - 10) = 4x - 25, 3x + 5 = 4x - 25. Check: 4(30) - 25 = 95 = 45 + 50.
- x = 12: the remote angles are 60^\circ and 42^\circ and the exterior angle is 102^\circ. From 5x + (3x + 6) = 10x - 18, 8x + 6 = 10x - 18, so 2x = 24. Check: 10(12) - 18 = 102 = 60 + 42.
- y = 12: the remote angles are 32^\circ and 40^\circ and the exterior angle is 72^\circ. From (2y + 8) + (3y + 4) = 7y - 12, 5y + 12 = 7y - 12, so 2y = 24.
- x = 9 only: the remote angles are 36^\circ and 45^\circ and the exterior angle is 81^\circ. From x^2 = 4x + (3x + 18), x^2 - 7x - 18 = 0, and (x - 9)(x + 2) = 0. The root x = -2 makes the remote angle 4x equal to -8, which is not an angle, so it is rejected.
- x = 40, and the triangle’s angles are 90^\circ, 45^\circ and 45^\circ. The remote angles of the exterior angle are the right angle and the other acute angle, so 90 + (x + 5) = 3x + 15, which gives 2x = 80. The exterior angle is 135^\circ, so the interior angle beside it is 180^\circ - 135^\circ = 45^\circ.
- 42^\circ each. 2b + 96 = 180, so 2b = 84.
- 104^\circ. The other base angle is also 38^\circ, and 180 - 38 - 38 = 104.
- 70^\circ and 40^\circ, or 55^\circ and 55^\circ. If 70^\circ is a base angle, the other base angle is 70^\circ and the vertex angle is 180 - 140 = 40. If 70^\circ is the vertex angle, each base angle is (180 - 70) \div 2 = 55.
- 45^\circ and 45^\circ. The right angle must be the vertex angle, since two right base angles would leave nothing; the base angles share 90^\circ equally.
- m\angle ADB = 80^\circ, m\angle ADC = 100^\circ, m\angle DAC = 45^\circ. In \triangle ABD, 180 - 42 - 58 = 80. The linear pair gives 180 - 80 = 100. In \triangle ADC, 180 - 100 - 35 = 45. Check: 58 + 45 = 103 = 180 - 42 - 35.
- 95^\circ. The diagonal makes two triangles whose six angles together make up the four angles of the quadrilateral, so the four add to 2 \times 180^\circ = 360^\circ, and 360 - 85 - 110 - 70 = 95.
- m\angle AEB = 65^\circ and m\angle EDC = 70^\circ. In \triangle ABE, 180 - 50 - 65 = 65. \angle CED is vertical to \angle AEB, so it is also 65^\circ, and in \triangle CDE, 180 - 65 - 45 = 70.
- m\angle ADC = 60^\circ and m\angle DAC = 50^\circ. \angle ADC is exterior to \triangle ABD, with remote angles 35^\circ and 25^\circ. Then in \triangle ADC, m\angle DAC = 180^\circ - 60^\circ - 70^\circ = 50^\circ.
- 110^\circ. The base angles are equal and add to 180^\circ - 40^\circ = 140^\circ, so each is 70^\circ. \angle PRS is exterior at R with remote angles \angle P and \angle Q: 40^\circ + 70^\circ = 110^\circ.
- m\angle B = m\angle ACB = 65^\circ and m\angle A = 50^\circ. \angle ACB is the supplement of 115^\circ, which is 65^\circ, and \angle B equals it as the other base angle. Then the exterior angle theorem gives m\angle A + 65 = 115.
- The third is 120^\circ, and the triangle’s angles are 70^\circ, 50^\circ and 60^\circ. One exterior angle at each vertex adds to 360^\circ (Example 12), so the third is 360^\circ - 110^\circ - 130^\circ. Each interior angle is the supplement of the exterior angle at its vertex.
- No. Two right angles already add to 180^\circ, which would leave 0^\circ for the third angle, and an angle of a triangle must be more than 0^\circ.
- The three angles add to 180^\circ and the right angle is 90^\circ of that, so the other two add to 180^\circ - 90^\circ = 90^\circ.
- m\angle C = 180^\circ - m\angle A - m\angle B and m\angle F = 180^\circ - m\angle D - m\angle E. The right sides are equal, because the angles subtracted are equal, so m\angle C = m\angle F.
- If two angles were each 90^\circ or more, those two alone would add to at least 180^\circ, leaving nothing for the third angle. So at most one angle is 90^\circ or more, and the other two are acute.
- No. The measurements add to 182^\circ, but a protractor reading is only accurate to about a degree, so three readings can easily be off by 2^\circ in total. The theorem is proved, not measured: the argument with the parallel line holds for every triangle, and the 2^\circ is measuring error.
- One can be acute: an exterior angle is acute exactly when the interior angle beside it is obtuse. Two cannot: two acute exterior angles would need two obtuse interior angles, and those alone would add to more than 180^\circ.
- The exterior angle equals the sum of both remote angles. Each remote angle is more than 0^\circ, so the sum is larger than either one on its own.
- The interior angle beside the exterior angle is 180^\circ - 90^\circ = 90^\circ, so the triangle has a right angle. The other two angles are the remote angles, and they add to the exterior angle, 90^\circ, so they are complementary.
- Call each base angle b^\circ. By the exterior angle theorem m\angle CAE = 2b, so the bisector makes m\angle FAC = b = m\angle ACB. These two angles lie on opposite sides of \overline{AC}, between \overrightarrow{AF} and \overline{BC}, so they are alternate interior angles, and equal alternate interior angles make the lines parallel (proving lines parallel).
- No. The exterior angle at D is 180^\circ - m\angle D, and the angles of a quadrilateral add to 360^\circ, so m\angle D = 360^\circ - (m\angle A + m\angle B + m\angle C). Substituting, the exterior angle equals m\angle A + m\angle B + m\angle C - 180^\circ. The subtraction in the triangle proof worked because both totals were 180^\circ; here one of them is 360^\circ.