Lesson 4: Angles and Pairs of Angles | State Board · Class 7 · Mathematics

Interior, exterior and adjacent angles

From daily life

Open a laptop. The screen and the keyboard are the two arms of an angle, and the gap between them is its inside. The table, the wall and the rest of the room are outside it. Now cut a round cake into slices: two neighbouring slices share one cut and the centre, but each slice has its own piece of cake. Neighbours like these are adjacent angles.

Simple definition

The interior of an angle is the open space between its two arms. The exterior is the rest of the plane, leaving out the arms themselves. A point lying on an arm is in neither of them.

Adjacent angles: Two angles that sit side by side are adjacent: they have the same vertex, they share one arm, and they do not overlap each other.

In the figure: ∠AKB has the vertex K and the arms KA and KB. The points H and J lie between the arms, so they are in the interior. The points U, W and Z lie outside, so they are in the exterior. The point F is on the arm KA, so it is in neither.

Rule 1: Interior, exterior or on an arm? — Look at where the point is. Between the two arms means interior; outside both arms means exterior; on an arm means neither.

Easy: In the figure, H is between the arms KA and KB, so H is in the interior of ∠AKB.
Medium: F lies on the arm KA, so it is neither in the interior nor in the exterior.
Exam type: The point Y is on the line KB but behind K (on the other side of K from B). An arm is a ray that starts at K, so Y is not on the arm, and it is not between the arms either. Hence Y is in the exterior.

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Interior, exterior and adjacent angles

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In the figure: In (i), ∠XWY and ∠YWZ have the vertex W and the arm WY in common, and they lie on opposite sides of WY, so they are adjacent. In (ii), ∠XWZ and ∠XWY share the arm WX, but ∠XWY lies inside ∠XWZ; they overlap, so they are not adjacent.

Rule 2: Recognising adjacent angles — Check three things: the same vertex, one common arm, and no overlap. If even one of them is missing, the angles are not adjacent.

Easy: ∠XWY and ∠YWZ → vertex W, common arm WY, X and Z on opposite sides of WY; adjacent.
Medium: ∠XWZ and ∠XWY → common arm WX, but ∠XWY lies inside ∠XWZ; they overlap, so they are not adjacent.
Exam type: Four rays OA, OB, OC, OD start from O in this order, turning anticlockwise. ∠AOB and ∠COD have no common arm, so they are not adjacent. ∠AOC and ∠COD share the arm OC and lie on its opposite sides, so they are adjacent.

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Interior, exterior and adjacent angles

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Rule 3: Adding and subtracting adjacent angles — The whole angle is the sum of its adjacent parts. If the whole and one part are known, subtract to get the other part.

Easy: ∠AOB = 38° and ∠BOC = 27° are adjacent, so ∠AOC = 38 + 27 = 65°.
Medium: ∠AOC = 83° and ∠AOB = 29°, so ∠BOC = 83 − 29 = 54°.
Exam type: (2x + 4)° and (3x + 6)° are adjacent angles that together make 85°. Then 2x + 4 + 3x + 6 = 85, so 5x + 10 = 85, 5x = 75 and x = 15. The angles are 2 × 15 + 4 = 34° and 3 × 15 + 6 = 51° (34 + 51 = 85).

Remember

A point on an arm is neither in the interior nor in the exterior. Two angles are adjacent only when all three conditions hold: the same vertex, one common arm and no overlap.

Common mistake

Calling two angles adjacent just because they share an arm. ∠XWZ and ∠XWY share the arm WX, yet ∠XWY lies inside ∠XWZ, so they are not adjacent.

Shortcut

For adjacent angles, whole = part + part. To find a missing part, subtract the known part from the whole (83° − 29° = 54°).

Spot the difference

Case	Example	Verdict
Point between the arms	H	Interior
Point on an arm	F	Neither interior nor exterior
Point outside the arms	U	Exterior
Share an arm but overlap	∠XWZ and ∠XWY	Not adjacent

Practice questions

Easy

1A point lies on an arm of an angle. Is it in the interior or in the exterior?
2Draw ∠CDE. Mark one point in its interior, one in its exterior and one on the arm DE.
3∠POQ = 33° and ∠QOR = 46° are adjacent angles. Find ∠POR.

Medium

4The ray OD lies inside ∠AOB. Are ∠AOD and ∠AOB adjacent angles? Why?
5∠AOC = 96° and the ray OB lies inside it with ∠AOB = 39°. Find ∠BOC.
6Two adjacent angles together make 102°, and one is 18° bigger than the other. Find both angles.

Challenging

7The rays OA, OB, OC, OD are in this order. ∠AOB = 22°, ∠BOC = 35° and ∠COD = 41°. Find ∠AOD. Among ∠AOB, ∠BOC and ∠COD, name all the pairs of adjacent angles.
8∠AOB and ∠BOC are adjacent, ∠AOB is 14° more than ∠BOC and ∠AOC = 84°. Find ∠AOB and ∠BOC.

Show answers (try them yourself first)

1Neither; a point on an arm is not in the interior or in the exterior. 2) Check that the first point is between the arms, the second is outside both arms and the third is on the ray DE. 3) 33 + 46 = 79°.
4No. They share the arm OA, but ∠AOD lies inside ∠AOB, so they overlap. 5) 96 − 39 = 57°. 6) Take the 18° off: 102 − 18 = 84 and 84 ÷ 2 = 42. The angles are 42° and 42 + 18 = 60° (42 + 60 = 102).
7∠AOD = 22 + 35 + 41 = 98°. The adjacent pairs are ∠AOB with ∠BOC, and ∠BOC with ∠COD (∠AOB and ∠COD have no common arm).
8Take the 14° off: 84 − 14 = 70 and 70 ÷ 2 = 35, so ∠BOC = 35° and ∠AOB = 35 + 14 = 49° (49 + 35 = 84).

Complementary angles

From daily life

The corner of your notebook is a right angle, 90°. Draw a line from that corner into the page: the corner is split into two angles, for example 35° and 55°. The two parts always fill the corner exactly, so they add up to 90°. Two angles like these are complementary.

Simple definition

Two angles whose sum is 90° are complementary, and each one is the complement of the other. They may sit side by side or lie far apart.

In the figure: In (i) the angles 32° and 58° are side by side and together fill a right angle (the small square), since 32 + 58 = 90. In (ii) the angles 24° and 66° are drawn apart, but 24 + 66 = 90, so they are still complementary.

Rule 1: Is it a complementary pair? — Add the two angles. If the sum is exactly 90°, they are complementary.

Easy: 26° and 64° → 26 + 64 = 90, so they are complementary.
Medium: 48° and 52° → 48 + 52 = 100, which is not 90, so they are not complementary.
Exam type: From 19°, 24°, 34°, 56°, 66° and 71°, the complementary pairs are 19° and 71°, 24° and 66°, 34° and 56° (each pair adds up to 90°).

Rule 2: Finding the complement — The complement of an angle is 90° minus that angle.

Easy: The complement of 28° is 90 − 28 = 62°.
Medium: The complement of 73° is 90 − 73 = 17°.
Exam type: An angle is 4 times its complement. Take the complement as 1 part and the angle as 4 parts: 5 parts make 90°, so 1 part = 90 ÷ 5 = 18°. The complement is 18° and the angle is 4 × 18 = 72° (72 + 18 = 90).

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Complementary angles

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Rule 3: Complementary angles with unknowns — Add the two expressions, put the sum equal to 90 and solve.

Easy: Two complementary angles differ by 12°. Take the smaller angle as x: x + x + 12 = 90, so 2x = 78 and x = 39. The angles are 39° and 39 + 12 = 51°.
Medium: (3a − 4)° and (2a + 14)° are complementary. Then 3a − 4 + 2a + 14 = 90, so 5a + 10 = 90, 5a = 80 and a = 16. The angles are 3 × 16 − 4 = 44° and 2 × 16 + 14 = 46°.
Exam type: In a right triangle the two acute angles are complementary (the three angles add up to 180° and the right angle is 90°). If they differ by 18°, the smaller angle x satisfies x + x + 18 = 90, so 2x = 72 and x = 36. The acute angles are 36° and 54°.

Remember

Complementary angles add up to 90°, one right angle, so each of them is acute. They do not have to be side by side; they may lie far apart.

Common mistake

Subtracting from 180° instead of 90°. The complement of 28° is 90 − 28 = 62°, not 180 − 28 = 152°.

Shortcut

After finding a complement, add the two angles back: they must make exactly 90° (28 + 62 = 90). This catches most slips.

Spot the difference

Angles	Sum	Complementary?
26° and 64°	90°	Yes
48° and 52°	100°	No
24° and 66°	90°	Yes
112° and 68°	180°	No (they are supplementary)

Practice questions

Easy

1Find the complement of 17°.
2Find the complement of 82°.
3Are 29° and 61° complementary angles?

Medium

4One acute angle of a right triangle is 41°. Find the other acute angle.
5(2x + 8)° and (x + 22)° are complementary. Find x and both angles.
6An angle is 24° less than its complement. Find the angle and its complement.

Challenging

7An angle is 8 times its complement. Find both angles.
8Two complementary angles are in the ratio 3 : 7. Find both angles.

Show answers (try them yourself first)

190 − 17 = 73°. 2) 90 − 82 = 8°. 3) 29 + 61 = 90, so yes, they are complementary.
490 − 41 = 49°. 5) 2x + 8 + x + 22 = 90, so 3x + 30 = 90, 3x = 60 and x = 20. The angles are 2 × 20 + 8 = 48° and 20 + 22 = 42° (48 + 42 = 90). 6) The two angles add up to 90° and differ by 24°. Take the 24° off: 90 − 24 = 66 and 66 ÷ 2 = 33. The angle is 33° and its complement is 33 + 24 = 57°.
7Take the complement as 1 part and the angle as 8 parts: 9 parts = 90°, so 1 part = 10°. The complement is 10° and the angle is 8 × 10 = 80°.
83 + 7 = 10 parts = 90°, so 1 part = 9°. The angles are 3 × 9 = 27° and 7 × 9 = 63° (27 + 63 = 90).

Supplementary angles

From daily life

A pencil stands on the straight edge of a ruler lying flat. It makes an angle on its left and another angle on its right, and together they fill the whole straight edge. If the left angle is 115°, the right one must be 180 − 115 = 65°. Two angles like these are supplementary.

Simple definition

Two angles whose sum is 180° are supplementary, and each one is the supplement of the other. They may sit side by side on a line or lie far apart.

In the figure: In (i) the angles 118° and 62° sit side by side on a straight line, and 118 + 62 = 180. In (ii) the angles 77° and 103° are drawn apart, but 77 + 103 = 180, so they are still supplementary.

Rule 1: Is it a supplementary pair? — Add the two angles. If the sum is exactly 180°, they are supplementary.

Easy: 112° and 68° → 112 + 68 = 180, so they are supplementary.
Medium: Two acute angles can never be supplementary, because each is less than 90°, so their sum is less than 180°. Two obtuse angles can never be supplementary either, because each is more than 90°.
Exam type: From 33°, 54°, 78°, 102°, 126° and 147°, the supplementary pairs are 33° and 147°, 54° and 126°, 78° and 102° (each pair adds up to 180°).

Rule 2: Finding the supplement — The supplement of an angle is 180° minus that angle.

Easy: The supplement of 62° is 180 − 62 = 118°.
Medium: The supplement of 149° is 180 − 149 = 31°.
Exam type: An angle is 28° more than its supplement. The two angles add up to 180° and differ by 28°. Take the 28° off: 180 − 28 = 152 and 152 ÷ 2 = 76. The supplement is 76° and the angle is 76 + 28 = 104° (104 + 76 = 180).

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Supplementary angles

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Rule 3: Supplementary angles with unknowns — Add the two expressions, put the sum equal to 180 and solve.

Easy: (x + 24)° and (x + 36)° are supplementary. Then x + 24 + x + 36 = 180, so 2x + 60 = 180, 2x = 120 and x = 60. The angles are 60 + 24 = 84° and 60 + 36 = 96°.
Medium: (4a − 10)° and (a + 20)° are supplementary. Then 4a − 10 + a + 20 = 180, so 5a + 10 = 180, 5a = 170 and a = 34. The angles are 4 × 34 − 10 = 126° and 34 + 20 = 54°.
Exam type: Two supplementary angles are in the ratio 4 : 11. Together they have 4 + 11 = 15 parts, and 15 parts = 180°, so 1 part = 12°. The angles are 4 × 12 = 48° and 11 × 12 = 132° (48 + 132 = 180).

Remember

Supplementary angles add up to 180°, a straight line. Two right angles are supplementary (90 + 90 = 180). If two supplementary angles are not right angles, one of them is acute and the other is obtuse.

Common mistake

Subtracting from 90° instead of 180°. The supplement of 62° is 180 − 62 = 118°, not 90 − 62 = 28°.

Shortcut

The supplement is always 90° more than the complement. For 28°: the complement is 62° and the supplement is 62 + 90 = 152°. Memory aid: C for Corner (90°), S for Straight (180°).

Spot the difference

Angles	Sum	Type
118° and 62°	180°	Supplementary
77° and 103°	180°	Supplementary
28° and 62°	90°	Complementary
100° and 70°	170°	Neither

Practice questions

Easy

1Find the supplement of 74°.
2Find the supplement of 128°.
3Are 59° and 121° supplementary angles?

Medium

4(3x − 12)° and (2x + 17)° are supplementary. Find x and both angles.
5An angle is 34° less than its supplement. Find both angles.
6The supplement of an angle is 4 times the angle. Find both angles.

Challenging

7Two supplementary angles differ by 52°. Find both angles.
8The supplement of an angle is 12° more than twice its complement. Find the angle.

Show answers (try them yourself first)

1180 − 74 = 106°. 2) 180 − 128 = 52°. 3) 59 + 121 = 180, so yes, they are supplementary.
43x − 12 + 2x + 17 = 180, so 5x + 5 = 180, 5x = 175 and x = 35. The angles are 3 × 35 − 12 = 93° and 2 × 35 + 17 = 87° (93 + 87 = 180). 5) Take the 34° off: 180 − 34 = 146 and 146 ÷ 2 = 73. The angle is 73° and its supplement is 73 + 34 = 107°. 6) Take the angle as 1 part and its supplement as 4 parts: 5 parts = 180°, so 1 part = 36°. The angle is 36° and its supplement is 4 × 36 = 144°.
7Take the 52° off: 180 − 52 = 128 and 128 ÷ 2 = 64. The angles are 64° and 64 + 52 = 116° (64 + 116 = 180).
8Call the angle x. Its supplement is 180 − x and its complement is 90 − x. So 180 − x = 2 × (90 − x) + 12 = 192 − 2x, which gives x = 12. The angle is 12° (check: the supplement is 168°, the complement is 78° and 2 × 78 + 12 = 168).

Opposite rays, linear pair and vertically opposite angles

From daily life

Two friends start from the same bus stop and walk in exactly opposite directions along a straight road. Their paths are opposite rays, and the angle between them is a straight angle, 180°. A pair of scissors shows another idea: when the scissors are open, the angle between the two blades is always the same as the angle between the two handles beyond the screw. Such angles are vertically opposite.

Simple definition

Two rays that start from the same point and go in exactly opposite directions are opposite rays; together they form a straight line. Two adjacent angles whose outer arms are opposite rays form a linear pair. When two straight lines cross, the angles that face each other are vertically opposite angles.

In the figure: In (i) the rays OP and OQ make an angle of 145°, so they are not opposite. In (ii) the angle is 180° and P, O, Q lie on one straight line, so OP and OQ are opposite rays.

Rule 1: Opposite rays — Two rays with a common starting point are opposite exactly when the angle between them is 180°, that is, when they lie along one straight line.

Easy: K, M, N lie on one straight line with M in the middle. The rays MK and MN start from M and go in opposite directions, so they are opposite rays.
Medium: ∠AOB = 175°. The rays OA and OB make 175°, not 180°, so they are not opposite rays.
Exam type: The points A, B, C, D lie on one line in this order. The rays BA and BC are opposite rays. The rays BC and BD point the same way (they are the same ray), so they are not opposite.

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Opposite rays, linear pair and vertically opposite angles

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In the figure: ∠AOC = 138° and ∠COB = 42° are adjacent, and their outer arms OA and OB are opposite rays. So they form a linear pair, and 138 + 42 = 180.

Rule 2: Linear pair — The two angles of a linear pair always add up to 180°, so each is the supplement of the other.

Easy: One angle of a linear pair is 127°, so the other is 180 − 127 = 53°.
Medium: (x + 14)° and (2x + 10)° form a linear pair. Then x + 14 + 2x + 10 = 180, so 3x + 24 = 180, 3x = 156 and x = 52. The angles are 52 + 14 = 66° and 2 × 52 + 10 = 114° (66 + 114 = 180).
Exam type: One angle of a linear pair is 9 times the other. Together they have 1 + 9 = 10 parts, and 10 parts = 180°, so 1 part = 18°. The angles are 18° and 9 × 18 = 162°.

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Opposite rays, linear pair and vertically opposite angles

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In the figure: Two lines cross at O. The two angles of 66° face each other, and the two angles of 114° face each other. Each 66° angle and its neighbour of 114° make 180°.

Rule 3: Vertically opposite angles — When two lines cross, the angles that face each other are equal. Two neighbouring angles form a linear pair, so the neighbours add up to 180°.

Easy: Two lines cross and one angle is 72°. The angle facing it is also 72°, and each of the other two angles is 180 − 72 = 108°.
Medium: (3x + 8)° and (5x − 22)° are vertically opposite, so they are equal: 3x + 8 = 5x − 22, so 30 = 2x and x = 15. Each angle is 3 × 15 + 8 = 53° (and 5 × 15 − 22 = 53°).
Exam type: Two lines cross, and two neighbouring angles are (2y + 20)° and (y + 10)°. Neighbours add up to 180°: 2y + 20 + y + 10 = 180, so 3y + 30 = 180, 3y = 150 and y = 50. The angles are 2 × 50 + 20 = 120° and 50 + 10 = 60°; the other two angles are again 120° and 60° (120 + 60 = 180).

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Opposite rays, linear pair and vertically opposite angles

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Remember

A linear pair is always supplementary, but supplementary angles are not always a linear pair: 77° and 103° add up to 180° even when they lie far apart. In a linear pair the two angles are side by side on a line.

Common mistake

Thinking that all four angles at a crossing are equal. Only the angles that face each other are equal (66° with 66°, 114° with 114°); the neighbours add up to 180°.

Shortcut

At a crossing of two lines, one known angle gives all four: the angle facing it is the same, and the two neighbours are each 180° minus it. For 35° the four angles are 35°, 145°, 35° and 145°.

Spot the difference

Pair	Position	Relation
Linear pair	Side by side on a straight line	Sum = 180°
Vertically opposite angles	Face to face at a crossing	Equal
Complementary angles	Together or apart	Sum = 90°
Supplementary angles	Together or apart	Sum = 180°

Practice questions

Easy

1One angle of a linear pair is 94°. Find the other angle.
2Two lines cross and one of the four angles is 38°. Find the other three angles.
3If ∠AOB = 180°, what can you say about the rays OA and OB?

Medium

4(3x + 7)° and (2x + 23)° form a linear pair. Find x and both angles.
5Two vertically opposite angles are (4x − 18)° and (2x + 20)°. Find x and the angles.
6The two angles of a linear pair differ by 24°. Find both angles.

Challenging

7Two lines cross. Two neighbouring angles are in the ratio 8 : 7. Find all four angles.
8The lines AB and CD cross at O. ∠AOC = (6x − 5)° and ∠BOD = (4x + 17)°. Find x, ∠AOC and ∠AOD.

Show answers (try them yourself first)

1180 − 94 = 86°. 2) The angle facing it is 38° and each of the other two is 180 − 38 = 142°. 3) OA and OB are opposite rays: A, O and B lie on one straight line.
43x + 7 + 2x + 23 = 180, so 5x + 30 = 180, 5x = 150 and x = 30. The angles are 3 × 30 + 7 = 97° and 2 × 30 + 23 = 83° (97 + 83 = 180). 5) The angles are equal: 4x − 18 = 2x + 20, so 2x = 38 and x = 19. Each angle is 4 × 19 − 18 = 58° (and 2 × 19 + 20 = 58°). 6) Take the 24° off: 180 − 24 = 156 and 156 ÷ 2 = 78. The angles are 78° and 78 + 24 = 102°.
78 + 7 = 15 parts = 180°, so 1 part = 12°. The neighbours are 8 × 12 = 96° and 7 × 12 = 84°, so the four angles are 96°, 84°, 96° and 84°.
8The angles ∠AOC and ∠BOD face each other, so they are equal: 6x − 5 = 4x + 17, so 2x = 22 and x = 11. ∠AOC = 6 × 11 − 5 = 61° (and 4 × 11 + 17 = 61°). ∠AOD is its neighbour, so ∠AOD = 180 − 61 = 119°.

Interior and exterior angles of polygons

From daily life

Cut a paper triangle of any shape and tear off its three corners. Place the three torn corners side by side with their points touching: they always fit exactly along a straight line, which is 180°. So the three angles of every triangle add up to 180°. A field with more sides can be cut into triangles in the same way.

Simple definition

The angles inside a closed figure, at its corners, are its interior angles. If you stretch one side of a triangle beyond a corner, the new angle formed outside, next to the interior angle of that corner, is an exterior angle.

Polygon rule: From one corner of a polygon, draw all the diagonals. This cuts the polygon into (sides − 2) triangles, so the sum of its interior angles is (sides − 2) × 180°.

Exterior angle property: A triangle's exterior angle equals the sum of the two interior angles that are far from it; these are its remote interior angles.

In the figure: In ΔABC the side BC is stretched to D. The new angle ∠ACD = 106° is an exterior angle. The interior angle next to it is ∠ACB = 74°, and 74 + 106 = 180. The two far interior angles are ∠A = 57° and ∠B = 49°, and 57 + 49 = 106, the same as the exterior angle.

Rule 1: Angle sum of a triangle — Add up the three angles inside any triangle and you always get 180°. If two angles are known, subtract their sum from 180° to get the third.

Easy: A triangle has angles 48° and 67°. The third angle is 180 − 48 − 67 = 65°.
Medium: The angles of a triangle are x°, (x + 12)° and (x + 24)°. Then x + x + 12 + x + 24 = 180, so 3x + 36 = 180, 3x = 144 and x = 48. The angles are 48°, 60° and 72° (48 + 60 + 72 = 180).
Exam type: An isosceles triangle has an apex angle of 44°. The two base angles are equal, and together they make 180 − 44 = 136°, so each base angle is 136 ÷ 2 = 68°.

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Interior and exterior angles of polygons

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In the figure: The heptagon ABCDEFG has 7 sides. The four dashed lines from the corner A cut it into 5 triangles, and each triangle has an angle sum of 180°. So the angles of the heptagon add up to 5 × 180 = 900°, which is (7 − 2) × 180°.

Rule 2: Angle sum of a polygon — A polygon can be cut into (sides − 2) triangles, so its interior angles add up to (sides − 2) × 180°.

Easy: A polygon with 9 sides has an angle sum of (9 − 2) × 180 = 7 × 180 = 1260°.
Medium: The angle sum of a polygon is 1620°. Then 1620 ÷ 180 = 9 is the number of sides minus 2, so the number of sides is 9 + 2 = 11.
Exam type: Four angles of a pentagon are 100°, 110°, 120° and 95°. The angle sum of a pentagon is (5 − 2) × 180 = 540°. The fifth angle is 540 − (100 + 110 + 120 + 95) = 540 − 425 = 115°.

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Interior and exterior angles of polygons

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Rule 3: Exterior angles — Stretch one side of a polygon: the angle formed outside, next to an interior angle, is an exterior angle. The interior angle and its exterior angle form a linear pair, so they add up to 180°.

Easy: The interior angle is 74°, so the exterior angle next to it is 180 − 74 = 106°.
Medium: The exterior angle is 132°, so the interior angle next to it is 180 − 132 = 48°.
Exam type: A triangle has angles 52°, 61° and 67°. The exterior angles are 180 − 52 = 128°, 180 − 61 = 119° and 180 − 67 = 113°. Every corner has two exterior angles (they face each other, so they are equal), hence a triangle has 6 exterior angles: 128°, 128°, 119°, 119°, 113° and 113°.

Rule 4: Exterior angle property — A triangle's exterior angle is as large as its two remote interior angles put together. The reason: the exterior angle plus the interior angle next to it make 180°, and the three interior angles also make 180°.

Easy: The remote interior angles are 38° and 64°, so the exterior angle is 38 + 64 = 102°.
Medium: An exterior angle is 128° and one remote interior angle is 79°. The other remote interior angle is 128 − 79 = 49°.
Exam type: The remote interior angles are (2x + 15)° and (x + 28)°, and the exterior angle is (5x − 11)°. Then 2x + 15 + x + 28 = 5x − 11, so 3x + 43 = 5x − 11, 54 = 2x and x = 27. The remote angles are 2 × 27 + 15 = 69° and 27 + 28 = 55°, and the exterior angle is 5 × 27 − 11 = 124° (69 + 55 = 124).

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Interior and exterior angles of polygons

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Remember

Triangle 180°, quadrilateral 360°, pentagon 540°, hexagon 720°: each extra side adds 180°. An exterior angle plus the interior angle next to it is always 180°.

Common mistake

Adding the wrong angles for the exterior angle property. The exterior angle equals the two far interior angles, not the one next to it. In the figure the exterior angle is 106°; the far angles are 57° and 49° (sum 106°), while the near angle 74° makes 180° with it.

Shortcut

To get a triangle's exterior angle, add the two far interior angles; there is no need to find the third angle first. To find the angle sum of a polygon, subtract 2 from the number of sides and multiply by 180° (7 sides: 5 × 180 = 900°).

Spot the difference

Feature	Interior angle ∠ACB	Exterior angle ∠ACD
Position	Inside the triangle	Outside, next to C
Size in the figure	74°	106°
How to find it	180° − 57° − 49°	57° + 49°

Practice questions

Easy

1Two angles of a triangle are 56° and 71°. Find the third angle.
2Find the angle sum of a polygon with 10 sides.
3A triangle has an exterior angle whose remote interior angles are 43° and 61°. Find that exterior angle.

Medium

4In a triangle, an exterior angle is 115° and one of its remote interior angles is 66°. Find the other remote interior angle and the interior angle next to the exterior angle.
5A polygon has an angle sum of 2160°. How many sides does it have?
6The angles of a quadrilateral are in the ratio 2 : 3 : 4 : 6. Find all four angles.

Challenging

7In ΔPQR the exterior angle at R is 118°, and ∠P is 18° more than ∠Q. Find ∠P, ∠Q and ∠R.
8A hexagon has four angles 120°, 130°, 110° and 124°, and the other two angles are equal. Find them.

Show answers (try them yourself first)

1180 − 56 − 71 = 53°. 2) (10 − 2) × 180 = 8 × 180 = 1440°. 3) 43 + 61 = 104°.
4The other remote angle is 115 − 66 = 49°. The interior angle next to the exterior angle is 180 − 115 = 65° (66 + 49 + 65 = 180). 5) 2160 ÷ 180 = 12, so the polygon has 12 + 2 = 14 sides. 6) 2 + 3 + 4 + 6 = 15 parts = 360°, so 1 part = 24°. The angles are 48°, 72°, 96° and 144° (48 + 72 + 96 + 144 = 360).
7The remote angles ∠P and ∠Q add up to 118°. Take the 18° off: 118 − 18 = 100 and 100 ÷ 2 = 50, so ∠Q = 50° and ∠P = 50 + 18 = 68°. ∠R = 180 − 118 = 62° (68 + 50 + 62 = 180).
8The angle sum of a hexagon is (6 − 2) × 180 = 720°. The four known angles add up to 120 + 130 + 110 + 124 = 484°. The remaining 720 − 484 = 236° is shared equally: 236 ÷ 2 = 118° each.