Vertical angles
Two lines cross. What does that force to be true about the four angles?
Where you are 8 in Angle types and pairs
Why this exists
Draw two straight lines through one point and you have made four angles without measuring anything. It looks like four separate facts to find out.
It is two. The angles come in matching pairs, and once you know one of them you know all four. That makes a crossing the cheapest information in geometry: it turns up inside triangles, inside circles, and under every pair of parallel lines cut by a transversal, and each time it hands you three angles for the price of one.
Most of the time that is how the fact gets used. A figure hands you one or two angle measures and asks for an angle that was never measured. You could put a protractor on the page, but a drawing is only a sketch, and the question is what the figure forces to be true, not what it happens to look like. Vertical angles are one link in that chain, and this lesson also covers the whole job of finding unknown angle measures, with every link named.
Definition
When two lines cross, they make four angles around the crossing point. Two of those angles are vertical angles when they sit opposite each other: they share the vertex and nothing else, and each side of one is the continuation of a side of the other.
In Figure 1, \angle 1 and \angle 3 are one pair of vertical angles, and \angle 2 and \angle 4 are the other.
Two angles that sit side by side at a crossing, such as \angle 1 and \angle 2, are not vertical. They share a whole side, and together they make a straight line. A pair like that is a linear pair, and its two measures add to 180^\circ.
The word “vertical” has nothing to do with up and down. It comes from vertex: the two angles share a vertex and nothing more.
Intuition
Hold one line still and turn the other slowly. The angle at the top opens, and the angle at the bottom opens by exactly as much, because it is the same turn seen from the other side of the crossing. The two angles on the left and the right close by that same amount.
Nothing here depends on the angle being a nice number or on the figure being symmetric. It works at 1^\circ and at 89^\circ and at every crossing you will ever draw.
Results
Vertical Angles Theorem. Vertical angles are congruent.
In measures, which is the form you use when you are hunting for a number:
The two sizes are tied to each other as well. Neighboring angles form a linear pair, so
One measurement at a crossing gives you all four.
Finding unknown angle measures
Finding an angle nobody measured comes down to a short list of facts, each about how angles sit together at one vertex:
- Parts add to the whole. When a ray lies inside an angle, the two smaller angles add to the big one. This is the angle addition postulate, and the facts below are built on it.
- A right angle is 90^\circ. Two angles whose measures add to 90^\circ are complementary. Two angles that sit side by side and fill a right angle are always complementary.
- A straight line is 180^\circ. Two angles whose measures add to 180^\circ are supplementary. A linear pair is always supplementary.
- A full turn is 360^\circ. The angles all the way round a point add to 360^\circ, and there is a reason it is that number.
- Vertical angles are equal. The theorem above.
Each fact turns a picture into an equation. Reach for them whenever the angle you want is connected to the angles you know by a chain of these facts, and especially when the measures are written as expressions in x, so that finding the angle means solving an equation first. The simpler case, one known total and one missing part, is finding a missing angle; a chase is what that becomes when the figure has more in it. Example 8 walks through the method one link at a time.
Why it is true
The proof uses nothing but linear pairs, twice.
\angle 1 and \angle 2 sit together along one straight line, so they form a linear pair:
\angle 2 and \angle 3 sit together along the other straight line, so they form a linear pair too:
Both sums equal 180^\circ, so they equal each other:
Subtract m\angle 2 from both sides:
The same argument with \angle 1 as the shared neighbor gives m\angle 2 = m\angle 4.
Notice what the proof leans on: only that a straight line measures 180^\circ. It never needs the lines to be at any particular slope, which is why the result holds for every crossing rather than for the ones that happen to look symmetric. This is Proposition 15 of Book I of Euclid’s Elements, and the argument there is essentially this one. Written out as a formal two-column proof, it is its own skill: proving vertical angles congruent.
Worked examples
Finding three angles from one
Two lines cross as in Figure 3. One angle measures 40^\circ. Find a, b and c.
The angle marked b is opposite the 40^\circ angle, so they are vertical angles and are equal:
The angle marked a sits next to the 40^\circ angle along one straight line. They form a linear pair, so they add to 180^\circ:
The angle marked c is opposite a, so c = a = 140^\circ.
Check. The four angles go all the way around the point, so they should add to 360^\circ: 40 + 140 + 40 + 140 = 360. They do.
Solving for a variable in a vertical pair
In Figure 4 the two marked angles are vertical angles. Find x, then find the measure of each of the four angles at the crossing.
Vertical angles are equal, so set the two expressions equal to each other and solve:
Substitute back to get the measure. Use both expressions, since they must agree:
Each marked angle measures 70^\circ. The other two angles each form a linear pair with a 70^\circ angle, so each measures 180^\circ - 70^\circ = 110^\circ.
The four angles are 70^\circ, 110^\circ, 70^\circ and 110^\circ.
Solving for a variable in a linear pair
Two lines cross. Two neighboring angles at the crossing measure (2y + 15)^\circ and (4y - 3)^\circ. Find y and all four angles.
SolutionThese two angles are side by side, not opposite, so they are not equal. They form a linear pair, which means they add to 180^\circ:
Then 2(28) + 15 = 71 and 4(28) - 3 = 109. The two angles measure 71^\circ and 109^\circ, and 71 + 109 = 180 as it should.
The angle opposite the 71^\circ angle is also 71^\circ, and the angle opposite the 109^\circ angle is also 109^\circ.
Deciding first which relationship you have, equal or adding to 180^\circ, is the whole problem. The algebra after that is routine.
A linear pair with one side known
In Figure 5, find x and the measure of the angle on the left.
The two angles sit side by side along one straight line, so they form a linear pair and add to 180^\circ:
The angle on the left measures 4(25) + 12 = 112, so it is 112^\circ.
Check. 112^\circ + 68^\circ = 180^\circ, and 112^\circ is obtuse, which is what the picture shows.
Three lines through one point
Three lines pass through one point, as in Figure 6. Find a, b, c and d.
With three lines there are six angles, and they come in three vertical pairs. Each pair is formed by the same two lines, sitting opposite each other.
The 40^\circ angle, the 65^\circ angle and a sit side by side above the horizontal line, filling the half of the picture on that side. Together they make a straight angle:
Now read off the vertical pairs. b is opposite the 40^\circ angle, c is opposite the 65^\circ angle, and d is opposite a:
Check. All six go once around the point: 2(40 + 65 + 75) = 2(180) = 360.
Two unknowns at one crossing
Two lines cross. One angle measures (5x)^\circ and the angle vertical to it measures (3x + 40)^\circ. An angle next to them measures (2y)^\circ. Find x and y.
SolutionWork with the vertical pair first, because it involves only x:
So each of that pair measures 5(20) = 100 degrees.
The angle (2y)^\circ is next to a 100^\circ angle, so the two form a linear pair:
So x = 20 and y = 40, and the four angles measure 100^\circ, 80^\circ, 100^\circ and 80^\circ.
When one angle is a right angle
Two lines cross, and one of the four angles is a right angle. Show that all four angles are right angles.
SolutionCall the right angle \angle 1, so m\angle 1 = 90^\circ, and number the rest around the point.
\angle 3 is vertical to \angle 1, so m\angle 3 = 90^\circ.
\angle 2 forms a linear pair with \angle 1, so m\angle 2 = 180^\circ - 90^\circ = 90^\circ.
\angle 4 is vertical to \angle 2, so m\angle 4 = 90^\circ.
All four are right angles. This is why lines that meet at one right angle are simply called perpendicular: there is no way to have one right angle at a crossing without having four.
Finding unknown angle measures: a chase through a right angle
In Figure 7, line AB has O on it, ray OE is perpendicular to AB (the small square says so), and line CD crosses AB at O. You know m\angle EOC = 40^\circ. Find x, the measure of \angle DOB.
Find a fact that touches what you know. The 40^\circ angle and x share no side and are not opposite each other, so no single fact joins them. Look instead for a fact that joins the 40^\circ angle to something. Ray OC lies inside the right angle \angle EOA, so \angle EOC and \angle COA fill a right angle between them.
Write it as an equation. Filling a right angle means the two measures add to 90^\circ:
so m\angle COA = 50^\circ. This is allowed because of the fact, not because of the picture: a square mark or the word “perpendicular” is what makes \angle EOA exactly 90^\circ.
Take the next link. \angle COA and \angle DOB are formed by the same two lines, CD and AB, and sit opposite each other at the crossing. They are vertical angles, so they are equal:
Check the whole figure, not just the answer. Going once around O, the five angles are \angle BOE = 90^\circ, \angle EOC = 40^\circ, \angle COA = 50^\circ, \angle AOD = 130^\circ (it forms a linear pair with the 50^\circ angle \angle DOB) and \angle DOB = 50^\circ. They add to 90 + 40 + 50 + 130 + 50 = 360, as a full turn must.
That is the whole method. Every step names the fact it uses, and the fact is what makes the step correct. When the measures are expressions, the equation takes longer to solve, but choosing the fact is still the real work, and the algebra after it is ordinary equation solving.
Splitting a right angle
\angle PQR is a right angle, and ray QS lies inside it. m\angle PQS = (2x + 4)^\circ and m\angle SQR = (3x - 9)^\circ. Find x and both angles.
SolutionQS lies inside the right angle, so the two parts fill it. By angle addition they add to the whole, 90^\circ, which makes them complementary:
Then m\angle PQS = 2(19) + 4 = 42^\circ and m\angle SQR = 3(19) - 9 = 48^\circ.
Check. 42 + 48 = 90.
Two angles joined through a third
Figure 8 has the same arrangement as Figure 7: OE \perp AB, and line CD crosses AB at O. Now m\angle EOC = (x + 20)^\circ and m\angle DOB = (2x + 10)^\circ. Find x and both angles.
The two marked angles are not neighbors and not vertical, so there is no fact that joins them directly. \angle COA touches both, so route the problem through it.
\angle COA and \angle DOB are vertical angles, so m\angle COA = (2x + 10)^\circ. Now \angle EOC and \angle COA fill the right angle \angle EOA, so they add to 90^\circ:
So m\angle EOC = 20 + 20 = 40^\circ and m\angle DOB = 2(20) + 10 = 50^\circ.
The equation says something worth noticing: in this arrangement, \angle EOC and \angle DOB are always complementary, even though they are on opposite sides of the figure. The vertical pair carries one of them round to sit next to the other.
Around a point
Four rays leave one point, as in Figure 9. Find x and the three marked angles.
The four angles go once around the point with no gaps and no overlaps, so they add to 360^\circ. The square marks the first one as 90^\circ:
The angles are 2(40) = 80^\circ, 3(40) + 10 = 130^\circ and 40 + 20 = 60^\circ.
Check. 90 + 80 + 130 + 60 = 360.
A complement and a supplement in one sentence
The supplement of an angle is 20^\circ more than three times its complement. Find the angle.
SolutionCall the angle a^\circ. Its complement is what it takes to make 90^\circ, which is (90 - a)^\circ, and its supplement is what it takes to make 180^\circ, which is (180 - a)^\circ. The sentence becomes an equation, word for word:
The angle is 55^\circ.
Check. Its complement is 35^\circ and its supplement is 125^\circ, and 3(35) + 20 = 125. An angle has a complement only if it is less than 90^\circ, and 55^\circ is.
The bisectors of a linear pair
Points A, O and B lie on a line, and ray OC rises from O. Ray OP bisects \angle AOC and ray OQ bisects \angle COB, as in Figure 10. Show that m\angle POQ = 90^\circ, wherever OC points.
No measure is given, so name the unknowns instead. A bisector splits an angle into two equal halves, so call the halves of \angle AOC each a, and the halves of \angle COB each b. Then m\angle AOC = 2a and m\angle COB = 2b.
\angle AOC and \angle COB form a linear pair, so
Ray OC lies inside \angle POQ, so by angle addition m\angle POQ = m\angle POC + m\angle COQ = a + b, and that is 90^\circ.
Nothing in the argument used where OC points, only that A, O and B are on a line. So the bisectors of any linear pair are perpendicular. This is the same chase used for a different purpose: not to find a number, but to prove that a number is forced.
Boundaries
The lines must be straight, and they must actually cross. Two rays that meet at their endpoints make one angle, not four, and nothing here applies. Two segments that would cross if they were extended do not form vertical angles until they are extended.
Only what the figure states counts. A line that looks straight, or an angle that looks square, gives you nothing until the figure marks it or the words say so. The 180^\circ and 90^\circ facts are about straight lines and right angles, not about ones that look close.
Vertical means opposite, not adjacent. \angle 1 and \angle 2 share a side and are supplementary, not equal. The equal angles are the ones sharing only the point.
With three or more lines, pairs are formed line by line. With three lines through one point, an angle’s vertical partner is the one formed by the same two lines. Two angles that merely look opposite across the figure but are formed by different pairs of lines need not be equal.
Some angles cannot be found. If two lines cross at O and a third ray from O is drawn with no measure attached, the angles it makes are not fixed: the ray can swing and every given measure stays true. When that happens, no amount of algebra will produce an answer, because the figure does not have one.
The answer has to be a real angle. A negative x is fine, but a negative or zero angle is not, and two complementary angles must each be less than 90^\circ. If solving gives an angle like that, the figure described cannot exist.
Congruent, not the same. \angle 1 \cong \angle 3 says their measures match. It does not say they are the same angle, and in a proof you still name which one you are using.
Where this goes next
Vertical angles are usually the first step in a chase rather than the answer. Given one angle at a crossing you can reach all four, and from there through a whole figure.
The facts in this lesson are about angles that share a vertex. The next facts connect angles at different vertices, and the chase works the same way with more links in it. In a triangle, the three angles add to 180^\circ, which is the triangle angle sum, and it quickly gives the exterior angle theorem and the angle sum of any polygon.
When the crossing lines are a transversal over two parallel lines, vertical angles are one of the relationships that let you carry an angle from one crossing to the other, and chasing angles across parallel lines is this same chase with one more fact on the list.
Inside a circle the same crossing does more work: two chords meeting make an angle equal to the average of the two arcs it cuts off, and that result starts from this one.
Example 13 points somewhere else again. Writing each step together with the fact that justifies it is already a proof, and proofs about angle pairs set exactly that out in full.
Exercises
Naming the pairs
Use Figure 1 for Exercises 1 to 3.
- Name the angle that is vertical to \angle 2.
- Name two different pairs of angles in Figure 1 that form a linear pair.
- True or false: \angle 1 and \angle 2 are vertical angles. Explain.
Finding measures
Two lines cross. One of the four angles has the measure given. Find the other three.
- 25^\circ
- 90^\circ
- 117^\circ
- Two lines cross, and one angle is three times the size of the angle next to it. Find all four angles.
- Two lines cross, and two neighboring angles differ by 30^\circ. Find all four angles.
- Segments \overline{AB} and \overline{CD} cross at E, and m\angle AEC = 38^\circ. Find m\angle BED and m\angle AED.
Using algebra
- A pair of vertical angles measure (4x - 12)^\circ and (2x + 30)^\circ. Find x and the measure of each angle.
- A pair of vertical angles measure (7x + 4)^\circ and (9x - 20)^\circ. Find x and the measure of each angle.
- Two neighboring angles at a crossing measure (x + 20)^\circ and (3x)^\circ. Find x and both angles.
- Two angles are complementary. One measures (4x - 6)^\circ and the other (2x)^\circ. Find x and both angles.
- Ray OB lies inside \angle AOC. m\angle AOB = (x + 12)^\circ, m\angle BOC = (2x - 3)^\circ and m\angle AOC = 99^\circ. Find x and both smaller angles.
- Five angles go once around a point, measuring x^\circ, (2x)^\circ, (3x)^\circ, (4x)^\circ and (5x)^\circ. Find all five.
- A pair of vertical angles measure (x^2)^\circ and (5x + 24)^\circ. Find every value of x that works, and the angle measure each one gives.
Three lines through one point
- Three lines pass through one point. Three neighboring angles, all on one side of one of the lines, are 50^\circ, 70^\circ and an unknown. Find all six angles.
- Three lines pass through one point and all six angles are equal. How large is each angle? Explain.
Two or more steps
- A pair of vertical angles measure (6x)^\circ and (4x + 36)^\circ. An angle next to them measures (y + 30)^\circ. Find x and y.
- In the arrangement of Figure 7, suppose m\angle EOC = 18^\circ instead. Find m\angle COA, m\angle DOB and m\angle AOD.
- In the arrangement of Figure 7, suppose m\angle EOC = (x + 6)^\circ and m\angle DOB = (3x)^\circ. Find x and both angles.
- Two lines cross. One of the four angles measures (2x + 10)^\circ, and the angle vertical to one of its neighbors measures (3x - 5)^\circ. Find x and all four angles.
- Points A, O and B lie on a line. Rays OC and OD are on the same side of it, in the order A, C, D, B. m\angle AOC = (x + 30)^\circ, m\angle COD = (2x)^\circ and m\angle DOB = (x + 10)^\circ. Find the three angles and m\angle COB.
- Two lines cross. Going around the crossing, the first three angles measure (4x)^\circ, (2y + 12)^\circ and (x + 51)^\circ. Find x, y and all four angles.
- Four angles go once around a point, and their measures are in the ratio 2 : 3 : 4 : 6. Find all four.
Complements and supplements in words
- An angle is 24^\circ more than its complement. Find the angle.
- The supplement of an angle is four times its complement. Find the angle.
- The supplement of an angle is 10^\circ less than five times its complement. Find the angle.
Reasoning
- Explain why a pair of vertical angles can never be one acute angle and one obtuse angle.
- Can the four angles at a crossing all have different measures? Explain.
- Two lines cross, and two neighboring angles at the crossing are equal. Show that the lines are perpendicular.
- The proof of the Vertical Angles Theorem uses only one fact about lines. Which fact is it, and why does that make the theorem true for every crossing?
- Can two angles be vertical angles and complementary at the same time? If so, find them.
- Show that, for any angle less than 90^\circ, its supplement is always 90^\circ more than its complement.
- Ray OC lies inside the right angle \angle AOB. Show that the bisectors of \angle AOC and \angle COB make a 45^\circ angle with each other, wherever OC is.
- Lines AB and CD cross at O and m\angle AOC = 70^\circ. Ray OE lies inside \angle AOC, with no measure given. Can m\angle EOB be found? Explain.
- Two angles are complementary, and they measure (x + 110)^\circ and (2x - 5)^\circ. Solve for x and explain what the answer tells you.
- Explain why one angle is enough to find all four angles where two lines cross, but one angle is not enough to find all six where three lines cross at one point.
Answers
- \angle 4. It sits opposite \angle 2 and shares only the crossing point with it.
- Any two of: \angle 1 and \angle 2, \angle 2 and \angle 3, \angle 3 and \angle 4, \angle 4 and \angle 1. Each pair shares a side and lies along one straight line.
- False. They share a whole side, so they are a linear pair, and their measures add to 180^\circ. Vertical angles share only the vertex.
- 25^\circ, 155^\circ, 155^\circ. The opposite angle is also 25^\circ; each neighbor is 180^\circ - 25^\circ = 155^\circ.
- All four are 90^\circ. The opposite angle is 90^\circ and each neighbor is 180^\circ - 90^\circ = 90^\circ.
- 117^\circ, 63^\circ, 63^\circ. The neighbors are 180^\circ - 117^\circ = 63^\circ.
- 45^\circ, 135^\circ, 45^\circ, 135^\circ. Neighbors form a linear pair, so x + 3x = 180, giving x = 45 and 3x = 135.
- 75^\circ, 105^\circ, 75^\circ, 105^\circ. From x + (x + 30) = 180, 2x = 150 and x = 75.
- m\angle BED = 38^\circ and m\angle AED = 142^\circ. \overrightarrow{EA} and \overrightarrow{EB} point in opposite directions, and so do \overrightarrow{EC} and \overrightarrow{ED}, so \angle BED is vertical to \angle AEC. \angle AED forms a linear pair with \angle AEC, so it is 180^\circ - 38^\circ.
- x = 21, and each angle is 72^\circ. Vertical angles are equal: 4x - 12 = 2x + 30, so 2x = 42. Check: 4(21) - 12 = 72 and 2(21) + 30 = 72.
- x = 12, and each angle is 88^\circ. From 7x + 4 = 9x - 20, 24 = 2x. Check: 7(12) + 4 = 88 and 9(12) - 20 = 88.
- x = 40, and the angles are 60^\circ and 120^\circ. Neighbors add to 180^\circ: (x + 20) + 3x = 180, so 4x = 160.
- x = 16, and the angles are 58^\circ and 32^\circ. Complementary angles add to 90^\circ: (4x - 6) + 2x = 90, so 6x = 96.
- x = 30, so m\angle AOB = 42^\circ and m\angle BOC = 57^\circ. The parts add to the whole: (x + 12) + (2x - 3) = 99, so 3x = 90.
- 24^\circ, 48^\circ, 72^\circ, 96^\circ and 120^\circ. Around a point: 15x = 360, so x = 24.
- x = 8, giving 64^\circ, or x = -3, giving 9^\circ. Setting the pair equal, x^2 = 5x + 24, so x^2 - 5x - 24 = 0 and (x - 8)(x + 3) = 0. Both values give a positive angle: 8^2 = 64 = 5(8) + 24, and (-3)^2 = 9 = 5(-3) + 24. A negative x is fine; only the angle has to be positive.
- 50^\circ, 70^\circ, 60^\circ, 50^\circ, 70^\circ, 60^\circ. The three on one side fill a straight angle, so the unknown is 180^\circ - 50^\circ - 70^\circ = 60^\circ. The other three are their vertical partners.
- 60^\circ. Any three neighboring angles lie along one straight line and fill 180^\circ, so each is 180^\circ \div 3 = 60^\circ. Equivalently, six equal angles fill 360^\circ.
- x = 18 and y = 42. From 6x = 4x + 36, x = 18, so the vertical pair are each 108^\circ. The neighbor is 180^\circ - 108^\circ = 72^\circ, so y + 30 = 72.
- m\angle COA = 72^\circ, m\angle DOB = 72^\circ, m\angle AOD = 108^\circ. \angle COA is the complement of 18^\circ, \angle DOB is vertical to \angle COA, and \angle AOD forms a linear pair with \angle DOB, so it is 180^\circ - 72^\circ.
- x = 21, so m\angle EOC = 27^\circ and m\angle DOB = 63^\circ. As in Example 10, the two are complementary: (x + 6) + 3x = 90, so 4x = 84.
- x = 35, and the angles are 80^\circ, 100^\circ, 80^\circ, 100^\circ. The angle vertical to a neighbor is the other neighbor, so the two angles form a linear pair: (2x + 10) + (3x - 5) = 180, so 5x = 175.
- 65^\circ, 70^\circ and 45^\circ, and m\angle COB = 115^\circ. The three fill a straight angle: (x + 30) + 2x + (x + 10) = 180, so 4x = 140 and x = 35. Then m\angle COB = 70^\circ + 45^\circ.
- x = 17 and y = 50; the angles are 68^\circ, 112^\circ, 68^\circ, 112^\circ. The first and third are vertical: 4x = x + 51, so x = 17 and each is 68^\circ. The second forms a linear pair with the first: 2y + 12 = 180 - 68 = 112, so y = 50.
- 48^\circ, 72^\circ, 96^\circ and 144^\circ. Write them as 2k, 3k, 4k, 6k; they add to 360^\circ, so 15k = 360 and k = 24.
- 57^\circ. With the angle a: a = (90 - a) + 24, so 2a = 114. Its complement is 33^\circ, and 57 - 33 = 24.
- 60^\circ. 180 - a = 4(90 - a), so 180 - a = 360 - 4a and 3a = 180. Check: the supplement 120^\circ is four times the complement 30^\circ.
- 65^\circ. 180 - a = 5(90 - a) - 10, so 180 - a = 440 - 5a and 4a = 260. Check: the complement is 25^\circ, and 5(25) - 10 = 115 = 180 - 65.
- Vertical angles are equal. An acute angle is less than 90^\circ and an obtuse angle is more than 90^\circ, so they cannot be equal.
- No. The angles come in two vertical pairs and each pair is equal, so there are at most two different measures.
- Call each of the equal neighbors x. They form a linear pair, so x + x = 180 and x = 90. One right angle at a crossing forces all four to be right angles (Example 7), so the lines are perpendicular.
- That a straight line makes a straight angle of 180^\circ, so neighboring angles along it add to 180^\circ. The proof never uses the slope of either line or any particular measure, only that fact, so it applies to every pair of crossing lines.
- Yes: both measure 45^\circ. Vertical angles are equal, so each is a; complementary means a + a = 90, so a = 45. Two lines crossing at 45^\circ give such a pair.
- For an angle a, (180 - a) - (90 - a) = 90, whatever a is. The supplement fills up to 180^\circ and the complement only to 90^\circ, so the supplement is always 90^\circ more.
- Call the halves of \angle AOC each a and the halves of \angle COB each b. The two angles fill the right angle, so 2a + 2b = 90 and a + b = 45. The angle between the bisectors is made of one half of each, a + b, so it is 45^\circ.
- No. \angle EOB forms a linear pair with \angle AOE, so m\angle EOB = 180^\circ - m\angle AOE, but m\angle AOE can be anything strictly between 0^\circ and 70^\circ as OE swings, so m\angle EOB can be anything strictly between 110^\circ and 180^\circ. The figure does not fix it.
- x = -5, which gives angles of 105^\circ and -15^\circ. From (x + 110) + (2x - 5) = 90, 3x = -15. No angle measures -15^\circ, and no angle of 105^\circ has a complement, so no such pair of angles exists.
- At a crossing, the angle you know forms a linear pair with each of its two neighbors and is vertical to the fourth, so every angle is tied to it. With three lines there are six angles in three vertical pairs, and the known angle settles only its own pair. The three angles on one side of a line add to 180^\circ, so the other two on that side share what is left, but nothing says how to split it. A second measure is needed.