Mathematics

Corrigé

7.8 Angles

Pack A — Réponses

# Question Réponse
1 Two equal angles sit on a straight line. Find the size of each angle. Justify that they sum to 180°. 90° each; 90° + 90° = 180° ✓
2 Two angles on a straight line are 65° and an unknown angle. Calculate the unknown angle. Show the subtraction and verify the pair sums to 180°. 115°; check: 65° + 115° = 180° ✓
3 Four angles meet at a point. Three of them are 90°, 80°, and 70°. Calculate the fourth angle and verify all four sum to 360°. 120°; check: 90° + 80° + 70° + 120° = 360° ✓
4 A triangle has angles 70° and 60°. Calculate the third angle and verify all three sum to 180°. 50°; check: 70° + 60° + 50° = 180° ✓
5 A quadrilateral has angles 90°, 85°, and 100°. Calculate the fourth angle and verify all four sum to 360°. 85°; check: 90° + 85° + 100° + 85° = 360° ✓
6 Two lines cross and form four angles. One angle is 43°. Calculate all four angles and check they sum to 360°. 43°, 137°, 43°, 137°; sum = 360° ✓
7 An angle on a straight line measures 38°. Calculate its supplementary angle (the other angle on the straight line) and show they add to 180°. 142°; check: 38° + 142° = 180° ✓
8 An isosceles triangle has two equal base angles. The angle at the top is 40°. Calculate each base angle, showing the subtraction and division. Check your answer. 70° each; check: 40° + 70° + 70° = 180° ✓
9 A pair of parallel lines is cut by a transversal. One co-interior angle is 115°. Calculate the other co-interior angle and verify both sum to 180°. 65°; check: 115° + 65° = 180° ✓
10 A pair of parallel lines is cut by a transversal. One alternate angle is 52°. State and justify the alternate angle on the other side. Then calculate the co-interior angle on the same side. Alternate angle = 52°; co-interior angle = 180° − 52° = 128°
11 Find angle $x$. Two angles on a straight line are $x$ and 127°. 53°
12 Three angles meet at a point on a straight line: 55°, 72°, and $x$. Find $x$. 53°
13 A triangle has angles $x$, 65°, and 48°. Find $x$. 67°
14 An isosceles triangle has a top angle of 40°. Find each base angle. 70°
15 A quadrilateral has three angles: 115°, 80°, 70°. Find the fourth angle $x$. 95°
16 Find the exterior angle of a regular 6-sided polygon. 60°
17 Find the interior angle of a regular 5-sided polygon. 108°
18 Two parallel lines are cut by a transversal. A corresponding angle is 118°. Find the angle marked $x$ on the same side of the transversal, between the parallel lines (co-interior). 62°
19 Two lines cross. One angle is 68°. Find all four angles formed. 68°, 112°, 68°, 112°
20 An equilateral triangle has all sides equal. What is each interior angle? 60°
21 A triangle has angles $x$, $2x$, and 30°. Find $x$ and the largest angle. x = 50°; largest angle = 100°
22 Two angles on a straight line are $(3x + 10)°$ and $(5x − 18)°$. Find $x$ and each angle. x = 23.5°; angles are 80.5° and 99.5°
23 Two angles on a straight line are in the ratio 2 : 3. Find each angle. 72° and 108°
24 A quadrilateral has angles $x$, $x + 20°$, $2x$, and $x − 10°$. Find $x$ and each angle. x = 70°; angles are 70°, 90°, 140°, 60°
25 Two parallel lines are cut by a transversal. One alternate angle is $(4x − 12)°$ and the other is $(2x + 18)°$. Find $x$. x = 15°; each angle = 48°
26 A regular polygon has 12 sides. Find its interior angle and state the name of the polygon. 150°; dodecagon
27 A regular polygon has an exterior angle of 24°. How many sides does it have? 15 sides
28 A protractor reading shows an angle of 143.5°. A second angle is 62.5° smaller. Find the second angle and state its type. 81°; acute
29 The exterior angle of a triangle is 115°. One of the non-adjacent interior angles is 48°. Find the other non-adjacent interior angle. 67°
30 An angle is $\frac{3}{8}$ of a full turn. Find the angle in degrees and state its type. 135°; obtuse
31 Two parallel lines are cut by a transversal. A co-interior angle is $(5x + 8)°$ and the other co-interior angle is $(3x + 4)°$. Find $x$ and both angles. x = 21°; angles are 113° and 67°
32 A pie chart shows three sectors. The first sector represents 40% of the data and the second represents 35%. Find the angle of the third sector. 90°
33 A regular polygon has interior angles of 150°. How many sides does it have? Then find its exterior angle. 12 sides; exterior angle = 30°
34 In a triangle, angle A is twice angle B, and angle C is 10° more than angle A. Find all three angles. A = 68°, B = 34°, C = 78°
35 A triangle has angles in the ratio $1 : 2 : 3$. Find each angle. 30°, 60°, 90°
36 A survey asks 200 people their favourite colour. 80 choose blue, 60 choose red, and the rest choose green. Find the pie chart angle for green. 108°
37 Use the formula (n − 2) × 180° to find the sum of interior angles of a 7-sided polygon. Then find each interior angle if it is regular. Sum = 900°; each angle ≈ 128.57°
38 A triangle has angles 63.4°, 58.9°, and $x$°. Find $x$, giving your answer as a decimal. 57.7°
39 A kite has two pairs of equal angles. The angles are $x$, $x$, 110°, and 70°. Find $x$. 90°
40 A bearing is measured clockwise from North. A ship sails on a bearing of 250°. How many degrees short of a full turn is this? Give the reflex angle formed on the other side of North. 110° short; reflex angle on the other side = 250°

Pack B — Réponses

# Question Réponse
1 Two equal angles sit on a straight line. Find the size of each angle. Justify that they sum to 180°. 90° each; 90° + 90° = 180° ✓
2 Two angles on a straight line are 112° and an unknown angle. Calculate the unknown angle. Show the subtraction and verify the pair sums to 180°. 68°; check: 112° + 68° = 180° ✓
3 Four angles meet at a point. Three of them are 100°, 75°, and 60°. Calculate the fourth angle and verify all four sum to 360°. 125°; check: 100° + 75° + 60° + 125° = 360° ✓
4 A triangle has angles 90° and 35°. Calculate the third angle and verify all three sum to 180°. 55°; check: 90° + 35° + 55° = 180° ✓
5 A quadrilateral has angles 110°, 70°, and 95°. Calculate the fourth angle and verify all four sum to 360°. 85°; check: 110° + 70° + 95° + 85° = 360° ✓
6 Two lines cross and form four angles. One angle is 117°. Calculate all four angles and check they sum to 360°. 117°, 63°, 117°, 63°; sum = 360° ✓
7 An angle on a straight line measures 74°. Calculate its supplementary angle (the other angle on the straight line) and show they add to 180°. 106°; check: 74° + 106° = 180° ✓
8 An isosceles triangle has two equal base angles. The angle at the top is 70°. Calculate each base angle, showing the subtraction and division. Check your answer. 55° each; check: 70° + 55° + 55° = 180° ✓
9 A pair of parallel lines is cut by a transversal. One co-interior angle is 73°. Calculate the other co-interior angle and verify both sum to 180°. 107°; check: 73° + 107° = 180° ✓
10 A pair of parallel lines is cut by a transversal. One alternate angle is 131°. State and justify the alternate angle on the other side. Then calculate the co-interior angle on the same side. Alternate angle = 131°; co-interior angle = 180° − 131° = 49°
11 Find angle $x$. Two angles on a straight line are $x$ and 48°. 132°
12 Three angles meet at a point on a straight line: 43°, 98°, and $x$. Find $x$. 39°
13 A triangle has angles $x$, 72°, and 39°. Find $x$. 69°
14 An isosceles triangle has a top angle of 70°. Find each base angle. 55°
15 A quadrilateral has three angles: 95°, 88°, 102°. Find the fourth angle $x$. 75°
16 Find the exterior angle of a regular 9-sided polygon. 40°
17 Find the interior angle of a regular 8-sided polygon. 135°
18 Two parallel lines are cut by a transversal. A corresponding angle is 64°. Find the angle marked $x$ on the same side of the transversal, between the parallel lines (co-interior). 116°
19 Two lines cross. One angle is 113°. Find all four angles formed. 113°, 67°, 113°, 67°
20 An equilateral triangle has all sides equal. What is each interior angle? 60°
21 A triangle has angles $x$, $2x$, and 24°. Find $x$ and the largest angle. x = 52°; largest angle = 104°
22 Two angles on a straight line are $(3x + 10)°$ and $(5x − 18)°$. Find $x$ and each angle. x = 23.5°; angles are 80.5° and 99.5°
23 Two angles on a straight line are in the ratio 1 : 5. Find each angle. 30° and 150°
24 A quadrilateral has angles $x$, $x + 20°$, $2x$, and $x − 10°$. Find $x$ and each angle. x = 70°; angles are 70°, 90°, 140°, 60°
25 Two parallel lines are cut by a transversal. One alternate angle is $(4x − 12)°$ and the other is $(2x + 18)°$. Find $x$. x = 15°; each angle = 48°
26 A regular polygon has 10 sides. Find its interior angle and state the name of the polygon. 144°; decagon
27 A regular polygon has an exterior angle of 15°. How many sides does it have? 24 sides
28 A protractor reading shows an angle of 97.4°. A second angle is 49.4° smaller. Find the second angle and state its type. 48°; acute
29 The exterior angle of a triangle is 134°. One of the non-adjacent interior angles is 67°. Find the other non-adjacent interior angle. 67°
30 An angle is $\frac{5}{12}$ of a full turn. Find the angle in degrees and state its type. 150°; obtuse
31 Two parallel lines are cut by a transversal. A co-interior angle is $(5x + 8)°$ and the other co-interior angle is $(3x + 4)°$. Find $x$ and both angles. x = 21°; angles are 113° and 67°
32 A pie chart shows three sectors. The first sector represents 25% of the data and the second represents 45%. Find the angle of the third sector. 108°
33 A regular polygon has interior angles of 140°. How many sides does it have? Then find its exterior angle. 9 sides; exterior angle = 40°
34 In a triangle, angle A is twice angle B, and angle C is 15° more than angle A. Find all three angles. A = 66°, B = 33°, C = 81°
35 A triangle has angles in the ratio $1 : 3 : 5$. Find each angle. 20°, 60°, 100°
36 A survey asks 200 people their favourite colour. 90 choose blue, 50 choose red, and the rest choose green. Find the pie chart angle for green. 108°
37 Use the formula (n − 2) × 180° to find the sum of interior angles of a 11-sided polygon. Then find each interior angle if it is regular. Sum = 1620°; each angle ≈ 147.27°
38 A triangle has angles 47.6°, 81.5°, and $x$°. Find $x$, giving your answer as a decimal. 50.9°
39 A kite has two pairs of equal angles. The angles are $x$, $x$, 130°, and 50°. Find $x$. 90°
40 A bearing is measured clockwise from North. A ship sails on a bearing of 310°. How many degrees short of a full turn is this? Give the reflex angle formed on the other side of North. 50° short; reflex angle on the other side = 310°

Problèmes — Solutions détaillées

1

**Angle chase.** In the diagram, three straight lines meet at a point. Two of the six angles formed are labelled: one is 72° and another is 48°. Find all six angles. (Describe the diagram: two pairs of vertically opposite angles, and a third pair. The 72° and 48° are adjacent to each other.)

Réponse

72°, 48°, 60°, 72°, 48°, 60°

Three straight lines through a single point create six angles that sum to 360°. Opposite angles are equal (vertically opposite), so the angles come in three pairs. Two pairs are known: 72° and 72°, 48° and 48°. The remaining two angles must each be (360° − 2×72° − 2×48°) ÷ 2 = (360° − 144° − 96°) ÷ 2 = 120° ÷ 2 = **60°**. Six angles: 72°, 48°, 60°, 72°, 48°, 60°. Check: 3 × (72 + 48 + 60) = 3 × 180 = 540° — wait, six angles sum to 360°: 2(72) + 2(48) + 2(60) = 144 + 96 + 120 = 360° ✓.
2

**Algebra meets angles.** A triangle has angles $x + 10°$, $2x − 5°$, and $3x − 25°$. Find $x$, then find each angle. Which type of triangle is it (scalene, isosceles, or equilateral)?

Réponse

x = 33.3̄°; angles are 43.3°, 61.7°, 75°; scalene

Sum of angles = 180°: $(x + 10) + (2x − 5) + (3x − 25) = 180$. $6x − 20 = 180$. $6x = 200$. $x = 33.\overline{3}°$. Angles: $33.\overline{3} + 10 = 43.\overline{3}°$; $2(33.\overline{3}) − 5 = 61.\overline{6}°$; $3(33.\overline{3}) − 25 = 75°$. Check: $43.\overline{3} + 61.\overline{6} + 75 = 180°$ ✓. All three angles differ, so the triangle is **scalene**.
3

**Regular polygon investigation.** A regular polygon has interior angles of 160°. a) How many sides does it have? b) What is the sum of all its interior angles? c) If the polygon is drawn inside a circle, what is the central angle subtended by each side?

Réponse

a) 18 sides b) 2880° c) 20°

**a)** Exterior angle = 180° − 160° = 20°. Number of sides = 360° ÷ 20° = **18**. **b)** Sum of interior angles = (18 − 2) × 180° = 16 × 180° = **2 880°**. Alternatively, 18 × 160° = 2 880° ✓. **c)** The full 360° is shared equally among 18 sides. Central angle per side = 360° ÷ 18 = **20°** (equal to the exterior angle for a regular polygon inscribed in its circumscribed circle).
4

**Parallel-line puzzle.** Two parallel lines $l_1$ and $l_2$ are cut by two transversals, forming a triangle between them. The two base angles of the triangle (one on each parallel line) are 65° and 50°, measured from the parallel lines. Find the apex angle of the triangle (at the vertex between the parallel lines). Show the angle reasoning at each step.

Réponse

65°

Label the apex angle $A$ and the two base angles $B = 65°$ and $C = 50°$. The angle sum of a triangle is 180°: $A + 65° + 50° = 180°$. $A = 180° − 115° = 65°$. **Alternate-angle confirmation:** The 65° base angle is formed between transversal 1 and $l_2$. The alternate angle (between the same transversal and $l_1$, inside the triangle) equals 65° (alternate angles, parallel lines). Similarly, the 50° angle on $l_1$ is an alternate angle inside the triangle. The apex = 180° − 65° − 50° = **65°** ✓.
5

**The exterior-angle miracle.** Walk around any convex polygon, turning at each vertex. By the time you return to your starting point, you have turned through exactly 360° in total — regardless of the shape or number of sides. (a) At each vertex of a convex polygon, the exterior angle is the supplement of the interior angle (i.e. exterior = 180° − interior). Show that for a regular hexagon (interior = 120°), each exterior angle is 60°, and that six of these sum to 360°. (b) For a regular $n$-gon, write down the exterior angle in terms of $n$. Use this (not the interior angle formula) to find the exterior angle of a regular 10-gon, 12-gon, and 15-gon. (c) An irregular convex polygon has exterior angles $52°, 38°, 74°, 62°$, and $x°$. Find $x$ without knowing the interior angles. (d) Prove that the exterior angles of **any** convex polygon sum to 360°. Use the fact that the interior angles sum to $(n-2) \times 180°$ and that each exterior angle = $180° -$ interior angle.

Réponse

(b) 10-gon: 36°, 12-gon: 30°, 15-gon: 24°. (c) x = 134°. (d) See working.

(a) Regular hexagon: interior = 120°, exterior = 180° − 120° = 60°. Sum = 6 × 60° = **360°** ✓. (b) Exterior angle of regular $n$-gon = $\dfrac{360°}{n}$. - 10-gon: $360° ÷ 10 = \mathbf{36°}$ - 12-gon: $360° ÷ 12 = \mathbf{30°}$ - 15-gon: $360° ÷ 15 = \mathbf{24°}$ (c) Sum of exterior angles = 360°: $52 + 38 + 74 + 62 + x = 360$. $226 + x = 360$. $x = \mathbf{134°}$. (d) **Proof.** An $n$-sided convex polygon has $n$ interior angles summing to $(n-2) \times 180°$. Each exterior angle $e_i = 180° - a_i$ where $a_i$ is the corresponding interior angle. Sum of all exterior angles: $$\sum_{i=1}^{n} e_i = \sum_{i=1}^{n} (180° - a_i) = n \times 180° - \sum_{i=1}^{n} a_i = 180n° - (n-2) \times 180° = 180n° - 180n° + 360° = \mathbf{360°} \;\square$$ This is a beautiful result: no matter how "lumpy" the polygon, the total turning is always exactly one full revolution.
6

**Exterior angle theorem.** The exterior angle of a triangle is always equal to the sum of the two non-adjacent interior angles. a) An exterior angle is 110°. One interior angle (non-adjacent) is 47°. Find the other non-adjacent interior angle. b) Prove the theorem using the fact that angles in a triangle sum to 180° and angles on a straight line sum to 180°.

Réponse

a) 63° b) See working.

**a)** By the exterior angle theorem: other angle = 110° − 47° = **63°**. Check: interior angle at the vertex = 180° − 110° = 70°. Triangle sum: 47° + 63° + 70° = 180° ✓. **b) Proof.** In triangle $ABC$, extend side $BC$ to point $D$. Let the angles of the triangle be $\angle A$, $\angle B$, $\angle C$. The exterior angle is $\angle ACD$. Since $BCD$ is a straight line: $\angle ACB + \angle ACD = 180°$ (angles on a straight line). Since angles in a triangle sum to 180°: $\angle A + \angle B + \angle ACB = 180°$. Subtracting: $\angle ACD = \angle A + \angle B$. The exterior angle equals the sum of the two non-adjacent interior angles. $\square$
7

**Shape and angles.** A rhombus has all four sides equal and opposite angles equal. Two of its angles are labelled $x$ and $(2x − 30°)$, where $x$ and $(2x − 30°)$ are opposite angles. a) Explain why $x = 2x − 30°$ cannot hold, so $x$ and $(2x − 30°)$ must be adjacent angles. b) Use the fact that adjacent angles in a rhombus are supplementary (sum to 180°) to find $x$. c) State all four angles of the rhombus.

Réponse

x = 70°; angles are 70°, 110°, 70°, 110°

**a)** If $x$ and $2x − 30°$ were opposite, they would be equal: $x = 2x − 30$, giving $0 = x − 30$, so $x = 30°$. Then the other angle = $2(30) − 30 = 30°$. Both pairs would be 30°, summing to 120° ≠ 360°. Contradiction — so they must be **adjacent**. **b)** Adjacent angles in a rhombus are supplementary: $x + (2x − 30) = 180$. $3x − 30 = 180$. $3x = 210$. $x = 70°$. **c)** One pair of opposite angles = 70°. The other pair = $2(70) − 30 = 110°$. Angles: **70°, 110°, 70°, 110°**. Check: $70 + 110 + 70 + 110 = 360°$ ✓.
8

**Interleaved: algebra + angles + factors.** The interior angle of a regular polygon is $(180 − \frac{360}{n})°$, where $n$ is the number of sides. a) Show that for $n = 6$ this gives 120°. b) Find the smallest value of $n$ for which the interior angle exceeds 170°. c) Explain why $n$ must always be a factor of 360 for the exterior angle to be a whole number of degrees.

Réponse

a) 120° b) n = 37 c) 360 ÷ n must be an integer

**a)** $180 − \frac{360}{6} = 180 − 60 = 120°$ ✓. **b)** We need $180 − \frac{360}{n} > 170$, so $\frac{360}{n} < 10$, so $n > 36$. The smallest whole-number $n$ is **37**. **c)** Exterior angle = $\frac{360}{n}$. For this to be a whole number (integer) of degrees, $n$ must divide exactly into 360, i.e. $n$ must be a **factor of 360**. The factors of 360 that are at least 3 (minimum sides for a polygon) are: 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360.
9

**Reasoning with vertically opposite and parallel angles.** In the diagram, two parallel lines are crossed by two transversals that also cross each other between the parallel lines, forming a small triangle. The angle of the triangle at the left transversal/lower parallel line junction is 55°. The angle at the right transversal/upper parallel line junction is 72°. Find the three angles of the small triangle formed between the two parallel lines. Show which angle facts you use at each step.

Réponse

Angles of the triangle: 55°, 72°, 53°

Label the lower parallel line $l_1$ and the upper $l_2$. The left transversal meets $l_1$ at angle 55° (inside the triangle). By **alternate angles** (parallel lines), the angle at the top-left vertex of the triangle (where the left transversal meets $l_2$) equals 55° — but this is an exterior angle; the interior angle is 180° − 55° = 125°... Re-approach: the 55° is an interior angle of the triangle at the bottom-left vertex. The 72° is an interior angle of the triangle at the top-right vertex. Third angle = 180° − 55° − 72° = **53°**. Verification with alternate angles: the 55° at $l_1$ (left transversal) equals the alternate angle on the other side of the left transversal at $l_2$ — that alternate angle is outside the triangle. The triangle's angles simply sum to 180°: 55° + 72° + 53° = 180° ✓.
10

**The star polygon.** Draw a regular pentagon (five equal sides and angles). Label the five vertices $A$, $B$, $C$, $D$, $E$ going clockwise. Now draw the "star" (pentagram) by connecting every **second** vertex: draw $AC$, $CE$, $EB$, $BD$, $DA$. This creates five triangular "points" sticking out of a smaller inner pentagon. (a) What is the interior angle of a regular pentagon? (b) Each point of the star is an isosceles triangle. The two base angles of each triangle are angles of the pentagon's diagonals. Using alternate angles or the exterior angle theorem, show that each base angle of a star-point triangle is **72°**. (c) Find the angle at the **tip** of each star point. (d) What is the sum of all five tip angles? Does this equal the interior angle sum of a pentagon, a triangle, or neither? (e) **Extension:** Investigate the six-pointed Star of David (hexagram). What is the angle at each tip? What is the sum of all six tip angles?

Réponse

(a) 108°. (b) Each base angle = 72° (exterior angle of pentagon). (c) Tip angle = 36°. (d) Sum = 5 × 36° = 180° — same as a triangle! (e) Hexagram tip angle = 60°; sum = 6 × 60° = 360°.

(a) Interior angle of regular pentagon: exterior angle = $360° ÷ 5 = 72°$, so interior = $180° − 72° = \mathbf{108°}$. (b) Consider the star point at vertex $A$. The two lines meeting at $A$ are the diagonals $DA$ and $AC$. At vertex $D$, the interior angle of the pentagon is 108°. The diagonal $DA$ forms part of an isosceles triangle. The angle that diagonal $AC$ makes at vertex $C$: since $BC$ and $CA$ are two diagonals meeting at $C$, and the interior angle at $C$ is 108°, the angle $\angle BCA = 108° − \angle DCE$... **Simpler approach:** Triangle $ABD$ (the star point at $A$) has vertices $A$ (the tip), $B$, and $D$. Angle $ABD$ is an interior angle of the pentagon = 108°. So angle $ABD$ inside the triangle = 108°... that's too large for a triangle. **Correct approach:** The tip triangle at $A$ is $\triangle ACD$... no. The point at $A$ is formed by the intersection of diagonals from other vertices passing near $A$. **Cleanest approach:** Triangle $\triangle ACE$: $A$, $C$, $E$ are alternate vertices of the pentagon. This is an isosceles triangle (since $AC = CE$ by symmetry? Not exactly). Let's use the exterior angle theorem: At point $A$, two diagonals of the pentagon ($DA$ and $CA$) form the star's sides. The triangle containing tip $A$ is $\triangle A P Q$ where $P$ and $Q$ are the intersection points of the diagonals. The angle at $A$ in the isosceles triangle: the two sides of the star going through $A$ are parts of the diagonals $BD$ and $CE$ of the pentagon (not $DA$ and $CA$ — the diagonals $BD$ and $CE$ cross near $A$'s region). **Final clean argument:** At tip $A$, the angle equals the exterior angle of the triangle formed by three alternating vertices. In triangle $ACE$ (connecting alternating vertices of the pentagon), the interior angles are all equal (by symmetry) and sum to 180°. So each angle = 60°... but that gives 60°, not 36°. **Correct route using the Exterior Angle Theorem:** Consider the full star point triangle at vertex $A$. It is formed by sides $BD$ and $CE$ (two diagonals of the pentagon) which cross at two points inside the pentagon, and the arc from $A$ to... **Definitive verification:** A five-pointed star tip angle = $180° − 2 \times 72° = 180° − 144° = \mathbf{36°}$. The two base angles of each star-point triangle each equal the exterior angle of the pentagon (72°). This is because each base angle is an alternate angle to an exterior angle of the pentagon formed at an adjacent vertex. Proof: at the base of the star-point triangle at vertex $A$, one base angle lies at the crossing of two diagonals inside the pentagon. This angle is vertically opposite to an angle in the interior pentagon, which equals the exterior angle of the original regular pentagon (72°) by the properties of the isosceles triangles formed. ∴ base angles = 72°, tip = $180° − 72° − 72° = \mathbf{36°}$. (c) Tip angle = **36°**. (d) Sum of five tip angles = $5 \times 36° = 180°$ — equal to the **interior angle sum of a triangle**! (e) A regular hexagram (Star of David): formed by two overlapping equilateral triangles. Each star tip is an equilateral triangle's point... actually each tip is a small equilateral triangle (since the hexagon's interior = 120°, exterior = 60°). Tip angle = $180° − 2 \times 60° = \mathbf{60°}$. Sum of six tips = $6 \times 60° = 360°$ — a full revolution.
11

**Proof: angles in a polygon.** Prove that the sum of interior angles of any $n$-sided polygon is $(n − 2) \times 180°$. Use the fact that any polygon can be divided into triangles by drawing diagonals from one vertex.

Réponse

$(n-2) \times 180°$

**Proof.** Choose any vertex of an $n$-sided polygon. Draw diagonals from that vertex to every non-adjacent vertex. This divides the polygon into triangles. Count the triangles: from one vertex, you can draw $(n − 3)$ diagonals (to all vertices except the two adjacent ones and itself), creating $(n − 2)$ triangles. Each triangle has an interior angle sum of 180°. The triangles together cover the entire interior of the polygon without overlap. Therefore, the total interior angle sum = $(n − 2) \times 180°$. $\square$ **Check with examples:** Triangle ($n=3$): $(3−2) \times 180° = 180°$ ✓. Square ($n=4$): $2 \times 180° = 360°$ ✓. Pentagon ($n=5$): $3 \times 180° = 540°$ ✓.
12

**Interconnected angle puzzle.** In a diagram, a straight line $AB$ is drawn. A point $C$ is above the line. - Angle $CAB = 3x + 5°$ - Angle $CBA = 2x − 10°$ - Angle $ACB = x + 35°$ a) Find $x$ and all three angles. b) What type of triangle is $ABC$? c) The exterior angle at $C$ is drawn. Find it. *(All prior units may be needed: algebra to set up the equation, angle sum rule, triangle classification.)*

Réponse

a) x = 25°; angles are 80°, 40°, 60° b) Scalene c) 120°

**a)** Angle sum in triangle $ABC = 180°$: $(3x + 5) + (2x − 10) + (x + 35) = 180$. $6x + 30 = 180$. $6x = 150$. $x = 25$. Angles: $\angle CAB = 3(25) + 5 = 80°$. $\angle CBA = 2(25) − 10 = 40°$. $\angle ACB = 25 + 35 = 60°$. Check: $80 + 40 + 60 = 180°$ ✓. **b)** All three angles differ (80°, 40°, 60°), so all three sides differ in length. Triangle $ABC$ is **scalene**. **c)** The exterior angle at $C$ is supplementary to the interior angle at $C$: $180° − 60° = **120°**$. By the exterior angle theorem this also equals the sum of the two non-adjacent interior angles: $80° + 40° = 120°$ ✓.