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Math 7 · 7.G.B.5

Finding unknown angle measures

A figure gives you some angles and asks for one that nobody measured. How do you get from the angles you know to the one you want?

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Vertical anglesMath 7Complementary and supplementary anglesMath 7
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Finding unknown angle measures

When you reach for it

A figure hands you one or two angle measures and asks for an angle that was never measured. A protractor on the page would only measure the drawing, and a drawing is a sketch. The question is what the figure forces to be true, not what it happens to look like.

What it forces comes from a short list of facts about angles that share a vertex. Each fact has its own lesson; here they are used together.

  • Parts add to the whole. When a ray lies inside an angle, the two smaller angles add to the big one. This is angle addition.
  • A right angle is 90^\circ. Two angles that together fill a right angle are complementary: their measures add to 90^\circ.
  • A straight line is 180^\circ. Two neighbors whose outer sides make a straight line are a linear pair, and their measures add to 180^\circ. More generally, angles that sit side by side and fill one side of a straight line add to 180^\circ.
  • A full turn is 360^\circ. The angles all the way round a point, with no gaps and no overlaps, add to 360^\circ, the number of Degrees in a full turn.
  • Vertical angles are equal. Where two lines cross, the angles opposite each other have equal measures. That is the vertical angles theorem.

Each fact turns part of a picture into an equation. Reach for this skill when the angle you want is joined to the angles you know by a chain of these facts, or when the measures are written as expressions in x, so that finding the angle means solving an equation first. One fact with plain numbers is finding a missing angle; this lesson is what that becomes when the figure has more in it.

The procedure

The lesson runs over two days. Part 1 works with numbers only: one fact at a time, then several in a row. Part 2 turns the same facts into equations when the measures are expressions in x. Examples 1 to 3 and the first two groups of exercises go with Part 1; Examples 4 and 5 and the remaining groups go with Part 2.

Part 1: one fact, then a chain of facts

Take Figure 1. Line AB has the point 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 AOD = 121^\circ and want x, the measure of \angle EOC.

x
121^\circ
 B E C A D O
Figure 1Three facts, used one after another, carry the 121^\circ round to x.

Look for a fact that touches what you know. The 121^\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 121^\circ angle to something. \angle AOD and \angle DOB share the side OD, and their other sides, OA and OB, make the straight line AB. They are a linear pair.

Write the fact as an equation. A linear pair adds to 180^\circ:

121^\circ + m\angle DOB = 180^\circ

Subtracting 121^\circ from both sides gives m\angle DOB = 59^\circ.

Take the next link. \angle DOB and \angle COA are formed by the same two lines, CD and AB, and sit opposite each other at the crossing. They are vertical angles, so m\angle COA = 59^\circ.

And the last. Ray OC lies inside the right angle \angle EOA, so \angle EOC and \angle COA fill it between them and are complementary:

x + 59^\circ = 90^\circ

so x = 31^\circ. This step rests on the square mark, not on the look of the picture: the mark is what makes \angle EOA exactly 90^\circ.

Check by another route. \angle COB is vertical to \angle AOD, so it is also 121^\circ. Ray OE lies inside it and splits it into the right angle \angle BOE and \angle EOC, so 90^\circ + 31^\circ should be 121^\circ, and it is.

Each link named the fact that allowed it. Choosing the next fact is the real work; the arithmetic after each choice is an addition or a subtraction.

Part 2: when the measures are expressions

A figure often labels an angle with an expression such as (4x + 6)^\circ. The angle is not yet a number, because x is not known, but every fact above still holds, and each fact now gives an equation in x.

Suppose \angle KLM is a right angle, ray LN lies inside it, m\angle KLN = (4x + 6)^\circ and m\angle NLM = 48^\circ.

Name the relationship. Ray LN splits the right angle into two parts, so the parts are complementary.

Write the equation the relationship gives. Complementary measures add to 90:

(4x + 6) + 48 = 90

Solve it. Add the numbers on the left, 6 + 48 = 54, to get 4x + 54 = 90. This is a two-step equation. Subtract 54 from both sides to get 4x = 36, then divide both sides by 4 to get x = 9.

Answer the question that was asked. x is not the angle. It is the number that, put into the expression, gives the angle: m\angle KLN = 4(9) + 6 = 42^\circ.

Check against the fact. 42^\circ + 48^\circ = 90^\circ, so the two parts really do fill the right angle.

In this lesson the unknown sits on one side of the equation only. When two angles in one equation both carry x, as in Example 5, the x-terms are combined first: x + 2x = 3x is the distributive property, (1 + 2)x, read backwards.

Worked examples

Example 1

Slices of a pizza

A round pizza is cut into five slices, and every cut runs to the center, as in Figure 2. Four of the slices have angles at the center of 76^\circ, 58^\circ, 83^\circ and 69^\circ. Find the angle x of the fifth slice.

76^\circ
58^\circ
83^\circ
69^\circ
 x
Figure 2Five slices meet at the center and go once round it.
Solution

The five slices meet at the center and go once round it, with no gaps and no overlaps, so their angles add to a full turn, 360^\circ:

76 + 58 + 83 + 69 + x = 360

The four known angles add to 286, so 286 + x = 360. Subtracting 286 from both sides gives x = 74. The fifth slice has an angle of 74^\circ at the center.

Check. 76 + 58 + 83 + 69 + 74 = 360.

Example 2

Two roads and a footpath

Two straight roads cross at O: one runs from A to B and the other from C to D. A straight footpath OE leaves the crossing between the roads toward A and C, as in Figure 3. m\angle AOE = 34^\circ and m\angle EOC = 49^\circ. Find x = m\angle BOD and m\angle COB.

34^\circ
49^\circ
 x B C E A D O
Figure 3Two roads and a footpath. The footpath has no partner on the far side.
Solution

The footpath OE lies inside \angle AOC, so by angle addition the two parts make the whole:

m\angle AOC = 34^\circ + 49^\circ = 83^\circ

\angle AOC and \angle BOD are both formed by the two roads, and they sit opposite each other at the crossing, so they are vertical angles and are equal: x = 83, that is, m\angle BOD = 83^\circ.

\angle COB and \angle BOD share the side OB, and their other sides, OC and OD, make the straight road CD. They are a linear pair, so m\angle COB = 180^\circ - 83^\circ = 97^\circ.

Nothing continues the footpath through O, so the first link had to be angle addition: only the whole of \angle AOC, made by the two roads, has a vertical partner.

Check. Going once round O: 34 + 49 + 97 + 83 + 97 = 360, where the last 97^\circ is \angle DOA, vertical to \angle COB.

Example 3

Two routes to one angle

Points A, O and B lie on a line, and ray OE is perpendicular to it at O. Ray OC lies between OA and OE, and ray OD lies between OE and OB, as in Figure 4. m\angle AOC = 64^\circ and m\angle COD = 81^\circ. Find x = m\angle DOB.

64^\circ
81^\circ
 x B D E C A O
Figure 4The marked 81^\circ angle straddles the perpendicular ray OE.
Solution

Route 1, through the right angles. OC lies inside the right angle \angle AOE, so m\angle COE = 90^\circ - 64^\circ = 26^\circ. Ray OE lies inside \angle COD, so by angle addition m\angle EOD = 81^\circ - 26^\circ = 55^\circ. Finally OD lies inside the right angle \angle EOB, so x = 90 - 55 = 35.

Route 2, along the line. \angle AOC, \angle COD and \angle DOB sit side by side and fill the straight angle \angle AOB, so they add to 180^\circ:

64 + 81 + x = 180

so 145 + x = 180 and x = 35.

Both routes give m\angle DOB = 35^\circ, as they must. Route 2 takes one link instead of three because it never uses the perpendicular at all. When a figure offers several facts, the one that ties the given angles straight to the wanted one saves work, and the other route becomes a free check.

Example 4

A crossing with one expression

Two straight roads cross. One of the four angles at the crossing measures (5x + 9)^\circ, the angle opposite it measures 79^\circ, and one of the other two angles is labeled (6x + 17)^\circ. Find x and all four angles.

Solution

Angles opposite each other where two lines cross are vertical angles, so they are equal. That gives an equation with the unknown on one side only:

5x + 9 = 79

Subtract 9 from both sides: 5x = 70. Divide both sides by 5: x = 14.

Now the third label can be evaluated: 6(14) + 17 = 84 + 17 = 101, so that angle is 101^\circ. It is a neighbor of the 79^\circ angle along one road, so the two must form a linear pair, and indeed 79 + 101 = 180. The fourth angle is opposite the 101^\circ one, so it is also 101^\circ.

The four angles are 79^\circ, 101^\circ, 79^\circ and 101^\circ. Had 6x + 17 come to anything other than 101, no such crossing could exist.

Example 5

The footpath, with expressions

Take the roads and footpath of Figure 3 again, with new measures: m\angle AOE = (x + 9)^\circ, m\angle EOC = (2x + 5)^\circ and m\angle COB = 109^\circ. Find x, both footpath angles and m\angle BOD.

Solution

The two expressions are the parts of \angle AOC, so the first job is to find that whole. \angle AOC and \angle COB share the side OC, and their other sides, OA and OB, make the straight road AB. They are a linear pair, so m\angle AOC = 180^\circ - 109^\circ = 71^\circ.

Now angle addition gives the equation:

\begin{aligned} (x + 9) + (2x + 5) &= 71 \\ 3x + 14 &= 71 \\ 3x &= 57 \\ x &= 19 \end{aligned}

The second line collects the x-terms, x + 2x = 3x, and the numbers, 9 + 5 = 14.

So m\angle AOE = 19 + 9 = 28^\circ and m\angle EOC = 2(19) + 5 = 43^\circ. Finally, \angle BOD is vertical to \angle AOC, so m\angle BOD = 71^\circ.

Check. 28 + 43 = 71, and going once round O: 28 + 43 + 109 + 71 + 109 = 360, where the last 109^\circ is \angle DOA, vertical to \angle COB.

Boundaries

Only what the figure states counts. A line that looks straight, or an angle that looks square, gives nothing until the figure marks it or the words say so. The 90^\circ and 180^\circ facts are about right angles and straight lines, not about ones that look close.

Some angles cannot be found. If a ray from the vertex is drawn with no measure attached and no fact pins it, the ray can swing and every given measure stays true. The angles it makes are then not fixed, and no amount of arithmetic or algebra will produce them, because the figure does not have an answer.

Vertical partners come from the same two lines. Two angles are vertical only when both are formed by the same pair of crossing lines. A lone ray, like the footpath in Example 2, has no vertical partner, and with three or more lines through a point an angle’s partner is the one made by the same two lines, not whichever angle looks opposite.

The answer has to be a real angle. A negative value of x is allowed, but every angle it gives must be greater than 0^\circ, the two angles of a complementary pair must each be less than 90^\circ, and the two of a linear pair each less than 180^\circ. If solving gives an angle that breaks this, the figure described cannot exist.

One vertex only. Every fact here is about angles that share a vertex. Angles at different corners of a figure, such as the three angles of a triangle, are tied together by other facts.

Where this goes next

The facts here all live at one point. The next facts join angles at different points, 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 leads to the angle sums of polygons. When a line crosses two parallel lines it carries angles from one crossing to the other, and parallel lines and transversals adds that fact to the list.

Writing each step together with the fact that allows it is already the shape of a proof. In the proof that vertical angles are equal that is set out in full, using the same linear-pair fact this lesson leans on.

Exercises

Part 1: one fact

  1. A ladder stands on flat, level ground and leans against a wall. On the side toward the wall, the ladder makes a 69^\circ angle with the ground. What angle does it make with the ground on the other side?
  2. A ceiling fan has five equally spaced blades. Find the angle between two neighboring blades, and the angle between two blades that have one blade between them, measured across that blade.
  3. A road runs east from a corner and a street runs north from the same corner, at a right angle to it. A straight path leaves the corner between them, making 23^\circ with the road. What angle does the path make with the street?
  4. Each blade of a pair of scissors runs straight on into its handle, so a blade and its handle make one straight line through the pivot. The blades are opened to 42^\circ. Find the angle between the two handles, and the angle between one blade and the other handle.
  5. Three straight paths leave a fountain. Seen from above, two of the angles between neighboring paths are 118^\circ and 135^\circ. Find the third.

Part 1: two facts in a row

  1. In the arrangement of Figure 1, suppose m\angle COB = 133^\circ instead. Find m\angle EOC, m\angle COA and m\angle DOB.
  2. Two searchlights stand at one spot on a straight shoreline and shine out over the sea. One beam makes 46^\circ with the shore toward the west, and the other makes 57^\circ with the shore toward the east. Find the angle between the two beams, and the angle between the eastern beam and the shore toward the west.
  3. Two lines cross, and one of the four angles measures 128^\circ. A ray from the crossing point splits one of the two smaller angles into two parts, one of which measures 19^\circ. Find the other part.
  4. Points A, O and B lie on a line. Ray OC is on one side of it with m\angle AOC = 131^\circ, and ray OD is on the other side with m\angle BOD = 65^\circ. Find m\angle COB and m\angle DOA, and check that the four angles round O add to 360^\circ.
  5. Lines AB and CD cross at O, and m\angle AOC = 38^\circ. Ray OE is perpendicular to line CD and lies inside \angle COB. Find m\angle COB and m\angle EOB.

Part 2: one relationship, one equation

  1. Two angles form a linear pair, measuring (4x + 11)^\circ and (2x - 5)^\circ. Find x and both angles.
  2. Two angles are complementary, measuring (3x - 8)^\circ and (x + 2)^\circ. Find x and both angles.
  3. Two lines cross. One angle measures (6x - 15)^\circ and the angle opposite it measures 81^\circ. Find x.
  4. Four angles go once round a point, measuring (x + 35)^\circ, (2x)^\circ, (3x - 11)^\circ and (2x + 16)^\circ. Find x and all four angles.
  5. Ray OB lies inside \angle AOC. m\angle AOB = (2x + 9)^\circ, m\angle BOC = 47^\circ and m\angle AOC = 104^\circ. Find x and m\angle AOB.
  6. A circle graph has five sectors. Three of them are equal, each with an angle of x^\circ at the center, and the other two have angles of 80^\circ and 64^\circ. Find x.
  7. Two angles are supplementary, and one of them is 24^\circ less than three times the other. Find both angles.

Part 2: two facts and an equation

  1. In the arrangement of Figure 1, suppose m\angle EOC = 23^\circ and m\angle AOD = (4x + 9)^\circ. Find x and m\angle AOD.
  2. Two lines cross. One of the four angles measures (3x + 14)^\circ, and its two neighbors together measure 200^\circ. Find x and all four angles.
  3. 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, and \angle COD is marked as a right angle. m\angle AOC = (x + 29)^\circ and m\angle DOB = (2x - 11)^\circ. Find x, both angles, and m\angle AOD.
  4. The scissors of Exercise 4 are opened again. Now the blades make an angle of (3x - 5)^\circ, and one blade makes (7x + 5)^\circ with the other handle. Find x and the angle between the two handles.

Reasoning

  1. 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, and m\angle COD = 50^\circ. Can m\angle AOC be found? What can be said about m\angle AOC + m\angle DOB? Explain.
  2. Four angles go once round a point, measuring (x + 150)^\circ, (x + 140)^\circ, (x + 120)^\circ and (2x - 10)^\circ. Solve for x and explain what the answer tells you.
  3. In Example 3, could the measures have been m\angle AOC = 64^\circ and m\angle COD = 120^\circ, with everything else as before? Explain.
  4. In Figure 1, explain why knowing any one of the five angles at O, other than the right angle \angle BOE, is enough to find all the others.
Answers
  1. 111^\circ. The ground is a straight line, so the two angles the ladder makes with it form a linear pair: 180 - 69 = 111.
  2. 72^\circ and 144^\circ. Five equal angles fill a full turn, so each is 360 \div 5 = 72, and by angle addition two of them make 2(72) = 144.
  3. 67^\circ. The road and the street make a right angle, and the path splits it into two complementary parts: 90 - 23 = 67.
  4. 42^\circ between the handles and 138^\circ between a blade and the other handle. The two blade-and-handle lines cross at the pivot, so the handles make the angle vertical to the blades’ angle, and a blade with the other handle forms a linear pair with it: 180 - 42 = 138.
  5. 107^\circ. The three angles go once round the fountain, so they add to 360^\circ: 360 - 118 - 135 = 107.
  6. m\angle EOC = 43^\circ, m\angle COA = 47^\circ, m\angle DOB = 47^\circ. Ray OE splits \angle COB into the right angle \angle BOE and \angle EOC, so 133 - 90 = 43. Then \angle COA fills the right angle \angle EOA with \angle EOC, 90 - 43 = 47 (it also forms a linear pair with \angle COB, and 180 - 133 = 47 agrees), and \angle DOB is vertical to \angle COA.
  7. 77^\circ between the beams and 123^\circ. The three angles fill one side of the straight shore: 46 + b + 57 = 180, so b = 77. The eastern beam and the shore toward the west form a linear pair with the 57^\circ angle: 180 - 57 = 123, which matches 46 + 77.
  8. 33^\circ. A neighbor of the 128^\circ angle forms a linear pair with it, so the smaller angles measure 180 - 128 = 52. The ray splits one of them, so the parts add to 52: 52 - 19 = 33.
  9. m\angle COB = 49^\circ and m\angle DOA = 115^\circ. Each forms a linear pair with the given angle on the same side of the line: 180 - 131 = 49 and 180 - 65 = 115. Round O: 131 + 49 + 65 + 115 = 360.
  10. m\angle COB = 142^\circ and m\angle EOB = 52^\circ. \angle COB forms a linear pair with \angle AOC: 180 - 38 = 142. Ray OE splits it into the right angle \angle COE and \angle EOB, so 142 - 90 = 52. Check: \angle BOD, vertical to \angle AOC, is 38^\circ, and 52 + 38 = 90.
  11. x = 29, and the angles are 127^\circ and 53^\circ. A linear pair adds to 180^\circ: (4x + 11) + (2x - 5) = 180, so 6x + 6 = 180 and 6x = 174.
  12. x = 24, and the angles are 64^\circ and 26^\circ. Complementary angles add to 90^\circ: (3x - 8) + (x + 2) = 90, so 4x - 6 = 90 and 4x = 96.
  13. x = 16. Vertical angles are equal: 6x - 15 = 81, so 6x = 96. Check: 6(16) - 15 = 81.
  14. x = 40, and the angles are 75^\circ, 80^\circ, 109^\circ and 96^\circ. Round a point: (x + 35) + 2x + (3x - 11) + (2x + 16) = 360, so 8x + 40 = 360 and 8x = 320.
  15. x = 24 and m\angle AOB = 57^\circ. The parts add to the whole: (2x + 9) + 47 = 104, so 2x + 56 = 104 and 2x = 48. Check: 57 + 47 = 104.
  16. x = 72. The five sectors fill a full turn: 3x + 80 + 64 = 360, so 3x + 144 = 360 and 3x = 216.
  17. 51^\circ and 129^\circ. Call the other angle a; the first is then 3a - 24, and supplementary angles add to 180: a + (3a - 24) = 180, so 4a = 204 and a = 51. The first is 3(51) - 24 = 129, and 51 + 129 = 180.
  18. x = 26 and m\angle AOD = 113^\circ. As in Part 1, run the chain the other way: m\angle COA = 90 - 23 = 67 (complementary), m\angle DOB = 67 (vertical), and \angle AOD forms a linear pair with \angle DOB, so it is 180 - 67 = 113. Then 4x + 9 = 113, so 4x = 104.
  19. x = 22, and the angles are 80^\circ, 100^\circ, 80^\circ, 100^\circ. The two neighbors of an angle at a crossing are opposite each other, so they are vertical and equal: each is 200 \div 2 = 100. Each forms a linear pair with the (3x + 14)^\circ angle, so 3x + 14 = 80 and 3x = 66.
  20. x = 24, so m\angle AOC = 53^\circ, m\angle DOB = 37^\circ and m\angle AOD = 143^\circ. The three angles fill one side of the line: (x + 29) + 90 + (2x - 11) = 180, so 3x + 108 = 180 and 3x = 72. Then m\angle AOD = 53 + 90 = 143 by angle addition.
  21. x = 18, and the handles make 49^\circ. The blades’ angle and the blade-to-other-handle angle form a linear pair: (3x - 5) + (7x + 5) = 180, so 10x = 180. The blades make 3(18) - 5 = 49^\circ (and the other angle 7(18) + 5 = 131^\circ; 49 + 131 = 180), and the handles make the angle vertical to the blades’, 49^\circ.
  22. m\angle AOC cannot be found, but the sum can: m\angle AOC + m\angle DOB = 130^\circ. The three angles fill one side of the line, so they add to 180^\circ, leaving 180 - 50 = 130 for the two outer ones. How that 130^\circ is shared is not fixed: OC and OD can turn together about O, keeping m\angle COD = 50^\circ and changing nothing given, so m\angle AOC can be any measure strictly between 0^\circ and 130^\circ.
  23. x = -8, which gives angles of 142^\circ, 132^\circ, 112^\circ and -26^\circ. From the full turn, 5x + 400 = 360, so 5x = -40. A negative x is allowed, but no angle measures -26^\circ, so no such figure exists: the first three angles alone already add to more than 360^\circ.
  24. No. \angle AOC, \angle COD and \angle DOB still fill one side of line AB, so 64 + 120 + x = 180, which gives x = -4. No angle measures -4^\circ: the two given angles alone cover 184^\circ, more than the straight angle they sit in.
  25. Every angle at O is tied to \angle COA by one fact: \angle EOC is its complement in the right angle \angle EOA, \angle DOB is vertical to it, and \angle AOD and \angle COB each form a linear pair with it. So any one of them gives m\angle COA, and m\angle COA gives the rest. The right angle is no help: it stays 90^\circ however line CD is turned.
kg1128 · updated 2026-10-07 · CC-BY-SA-4.0