The sum of all the exterior angles of a polygon is always 360^o

When you sum all the exterior angles, you should get 360^o

(2x + 5)⁰ + (2x)⁰ + (x - 20)⁰ + (x)⁰ + (3x + 10)⁰ + (x + 15)⁰ = 360⁰

2x + 5 + 2x + x - 20 + x + 3x + 10 + x + 15 = 360

10x + 10 = 360

10x = 360 - 10

10x = 350

Divide both sides by 10

x = $\frac{350}{10}$ = 35

Exterior angle = $\frac{360}{\mathrm{No\; of\; sides}}$

Let x = exterior angle

Interior angle = 2 x the exterior angle
Interior angle = 2 x x

Interior angle + Exterior angle = 180^o

2x + x = 180^o
3x = 180^o

Divide both sides by 3

x = $\frac{180}{3}$

x = 60^o

The exterior angle = 60^o

60 = $\frac{360}{\mathrm{No\; of\; sides}}$

60 x No of sides = 360

Divide both sides by 60

No of sides = $\frac{360}{60}$

No of sides = 6

Concept

Sum of exterior angles of any polygon = 360°

61° + 76° + (x + 10)° + (2x + 45)° = 360°
61 + 76 + x + 10 + 2x + 45 = 360
3x + 192 = 360
3x = 360 - 192
3x = 168

Divide both sides by 3

x = $\frac{168}{3}$

x = 56

n = k.

Sum of interior of regular polygon with k sides = (k - 2) x 180

But the sum is given as (3k - 10) x 90°

Note: right angle = 90°

(k - 2) x 180 = (3k - 10) x 90

Divide both sides by 90

(k - 2) x 2 = (3k - 10) x 1

2k - 4 = 3k - 10

10 - 4 = 3k - 2k

6 = k

∴ the polygon has 6 sides.

Number of sides of the polygon = 6

Exterior angle of the polygon = $\frac{\mathrm{360°}}{6}$ = 60°

Exterior angle of a regular polygon or not

Exterior angle of a regular polygon x n = 360°

n = $\frac{\mathrm{360°}}{\mathrm{Exterior\; angle\; of\; a\; regular\; polygon}}$

If n is whole number then the angle is an exterior angle.

66°

n = $\frac{\mathrm{360°}}{\mathrm{66°}}$ = 5.4545

Not an exterior angle of a regular polygon since n is not a whole number.

72°

n = $\frac{\mathrm{360°}}{\mathrm{72°}}$ = 5

Is an exterior angle of a regular polygon since n is a whole number.

24°

n = $\frac{\mathrm{360°}}{\mathrm{24°}}$ = 15

Is an exterior angle of a regular polygon since n is a whole number.

15°

n = $\frac{\mathrm{360°}}{\mathrm{15°}}$ = 24

Is an exterior angle of a regular polygon since n is a whole number.

168 = $\frac{\mathrm{(n\; -\; 2)\; x\; 180°}}{\mathrm{n}}$

168 x n = (n - 2) x 180

168n = 180n - 360

360 = 180n - 168n

360 = 12n

Divide both sides by 12.

n = $\frac{360}{12}$

n = 30

Number of sides of the polygon = 5 (since 5 angles are given)

Sum of angles = (5 - 2) x 180°

Sum of angles = 3 x 180°

Sum of angles = 540°

3x° + 2x° + 4x° + 3x° + 6x° = 540°

18x° = 540°

Divide both sides by 18

x = $\frac{540}{18}$ = 30

The smallest angle = 2x°

The smallest angle = 2 x 30°

The smallest angle = 60°

Pentagon has five sides.

n = 5

Sum of angles = (5 - 2) x 180°

Sum of angles = 3 x 180° = 540°

x° + (x + 5)° + (x + 10)° + (x + 15)° + (x + 20)° = 540°

x + x + 5 + x + 10 + x + 15 + x + 20 = 540

5x + 50 = 540

5x = 540 - 50

5x = 490

Divide both sides by 5

x = $\frac{490}{5}$ = 98

Pentagon has five sides hence n = 5.

Sum of the interior angles = (5 - 2) x 180°

Sum of the interior angles = 3 x 180°

Sum of the interior angles = 540°

y° + 2x° + 3x° + 2x° + y° = 540°

2y + 7x = 540 ---- eqn 1

But y = $\frac{\mathrm{3x}}{2}$

Make x the subject and substitute it into --- eqn 1

y = $\frac{\mathrm{3x}}{2}$

Get rid of the fraction by multiplying both sides by the denominator, 2

2 x y = 2 x $\frac{\mathrm{3x}}{2}$

2y = 3x

Divide both sides by 3.

x = $\frac{\mathrm{2y}}{3}$ --- eqn 2

Substitute --- eqn 2 into --- eqn 1

2y + 7 x $\frac{\mathrm{2y}}{3}$ = 540

Get rid of the fraction by multiplying both sides by the denominator, 3

3 x 2y + 3 x 7 x $\frac{\mathrm{2y}}{3}$ = 540 x 3

6y + 14y = 1,620

20y = 1,620

Divide both sides by 20

y = $\frac{1620}{20}$ = 81

Isosceles triangles (triangle with two sides equal) have the base angles the same.

∠ YXZ = ∠ XZY

Let ∠ YXZ = ∠ XZY = x

x + x + 40° = 180° (Sum of angles in a triangle)

2x = 180° - 40°

2x = 140°

Divide both sides by 2.

x = $\frac{\mathrm{140°}}{2}$ = 70°

∠ YXZ = ∠ XZY = 70°

∠ XZY + ∠XZT = 180° (Sum of angles on a straight line)

70° + ∠XZT = 180°

∠XZT = 180° - 70°

∠XZT = 110°

The midpoint divides XY into two equal lengths, 3 cm each.

tan(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Adjacent}}$

tan ∠ XZT = $\frac{\mathrm{XT}}{\mathrm{ZT}}$

XT = 3 cm but ZT is unknown.

You can use Pythagoream theorem to calculate ZT.

From pythagoras theorem

The longest side (side opposite/facing the right-angle) is the hypotenuse (c)

ZT² + XT² = XT²

ZT² + 3² = 6²

ZT² + 9 = 36

ZT² = 36 - 9

ZT² = 27

Take square root of both sides.

ZT = $\sqrt{\mathrm{27}}$

27 = 9 x 3

ZT = $\sqrt{\mathrm{9\; x\; 3}}$

But $\sqrt{\mathrm{a\; x\; b}}$ = $\sqrt{a}$ x $\sqrt{b}$

ZT = $\sqrt{9}$ x $\sqrt{3}$

$\sqrt{9}$ = 3

ZT = 3 x $\sqrt{3}$ = 3$\sqrt{3}$

tan ∠ XZT = $\frac{3\sqrt{3}}{3}$

3 cancel out 3.

tan ∠ XZT = $\sqrt{3}$

∠ QFG + ∠ EFG = 180°

105° + ∠ EFG = 180°

∠ EFG = 180° - 105°

∠ EFG = 75°

∠ FEG + ∠ EGF + ∠ EFG = 180° (Sum of angles of a triangle)

50° + m + 75° = 180°

m + 125° = 180°

m = 180° - 125°

m = 55°

n = ∠ FGS

∠ FGS = 180° - 105°(Corresponding angle to line RS)

∠ FGS = 75°

n = ∠ FGS = 75°

Pentagon has five sides.

n = 5

Sum of angles = (5 - 2) x 180°

Sum of angles = 3 x 180° = 540°

x° + (x + 5)° + (x + 10)° + (x + 15)° + (x + 20)° = 540°

x + x + 5 + x + 10 + x + 15 + x + 20 = 540

5x + 50 = 540

5x = 540 - 50

5x = 490

Divide both sides by 5

x = $\frac{490}{5}$ = 98

∠ PQR : ∠ QPS = 3 : 7

∠ PQR + ∠ QPS = 180° (Respective consecutive angles are supplementary)

Sum of ratios = 3 + 7 = 10

The sum of ratios represents the sum of the angles.

If 10 = 180°

3 = $\frac{\mathrm{3\; x\; 180°}}{10}$ = 54°

∴ ∠ PQR = 54°

If 10 = 180°

7 = $\frac{\mathrm{7\; x\; 180°}}{10}$ = 126°

∴ ∠ QPS = 126°

Alternatively,

∠ QPS = 180° - ∠ PQR

∠ QPS = 180° - 54° = 126°

∠ PSR = ∠ PQR = 54° (Opposite angles are equal)

∠ QSR = $\frac{\mathrm{\angle \; PSR}}{2}$ = $\frac{\mathrm{54°}}{2}$ = 27°