0580/21 Oct/Nov 2025 Worked Answers (IGCSE Maths 0580 Extended)

38 questions · 100 marks · 120 minutes

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1. Question 1

   2 marksRatio
   Step 1: Find the total number of parts: $2 + 3 = 5$. Step 2: Find the value of one part: $90 \div 5 = 18$. Step 3: Multiply by each ratio part: $2 \times 18 = 36$ and $3 \times 18 = 54$.
   Method:

   Total parts = $2 + 3 = 5$, so one part = $90 \div 5 = 18$. The amounts are $2 \times 18 = \$36$ and $3 \times 18 = \$54$.
   Examiner tips

   • Always check that the two amounts add up to the original total
   • Find the value of one part first by dividing the total by the sum of the ratio parts

2. Question 2

   3 marksAngle properties of triangles
   Step 1: Find angle $ACB$: angles at a point or on a straight line give $360 - 148 - 82 = 130°$. Step 2: Since $AC = BC$, triangle $ABC$ is isosceles, so base angles are equal. Step 3: $\text{Angle } CAB = (180 - 130) \div 2 = 25°$.
   Method:

   Find angle $ACB = 360 - 148 - 82 = 130°$. Then angle $CAB = (180 - 130) \div 2 = 25°$ using the isosceles property.
   Examiner tips

   • Look for isosceles triangles indicated by equal sides
   • Remember the angle sum of a triangle is $180°$

3. Question 3

   2 marksInterior angles of polygons
   Step 1: Exterior angle $= \frac{360}{20} = 18°$. Step 2: Interior angle $= 180 - 18 = 162°$. Alternatively: Interior angle sum $= 180(20 - 2) = 3240°$, so each interior angle $= \frac{3240}{20} = 162°$.
   Method:

   Exterior angle $= \frac{360}{20} = 18°$, so interior angle $= 180 - 18 = 162°$.
   Examiner tips

   • Remember that interior angle + exterior angle = $180°$
   • The sum of exterior angles of any polygon is $360°$

4. Question 4

   2 marksArea of a triangle
   Step 1: Use the area formula: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$. Step 2: Substitute: $12 = \frac{1}{2} \times 8 \times h$. Step 3: Solve: $12 = 4h$, so $h = 3$ cm.
   Method:

   $\frac{1}{2} \times 8 \times h = 12$, so $4h = 12$, giving $h = 3$ cm.
   Examiner tips

   • Always use the formula $\text{Area} = \frac{1}{2} \times b \times h$
   • Check your answer by substituting back into the formula

5. Question 5a

   2 marksTranslations
   Step 1: Identify corresponding points on triangles $T$ and $P$. Step 2: Find the change in $x$ and $y$: the triangle has moved $3$ units left and $1$ unit up. Step 3: The transformation is a translation by $\begin{pmatrix} -3 \\ 1 \end{pmatrix}$.
   Method:

   Compare corresponding vertices to find the translation vector $\begin{pmatrix} -3 \\ 1 \end{pmatrix}$.
   Examiner tips

   • Always state the type of transformation and then all required details
   • For translations, give the column vector

6. Question 5b

   2 marksEnlargement
   Step 1: For an enlargement, scale factor 2 from origin, multiply each coordinate by 2. Step 2: $(2, 1) \to (2 \times 2, 1 \times 2) = (4, 2)$.
   Method:

   Find vectors from $(3,3)$ to each vertex of $T$, multiply by $2$, then add back to $(3,3)$ to find the image vertices.
   Examiner tips

   • Always measure distances from the centre of enlargement, not from the origin
   • Check that each side of the image is the correct multiple of the original

7. Question 6a

   1 marksIndices
   Step 1: Use the index law $a^m \times a^n = a^{m+n}$. Step 2: $5^{-5} \times 5^5 = 5^{-5+5} = 5^0 = 1$.
   Examiner tips

   • Remember $a^0 = 1$ for any non-zero value of $a$

8. Question 6b

   2 marksIndices
   Step 1: The fractional index $\frac{2}{3}$ means cube root then square (or vice versa). Step 2: $\sqrt[3]{125} = 5$. Step 3: $5^2 = 25$.
   Method:

   $125^{\frac{2}{3}} = (\sqrt[3]{125})^2 = 5^2 = 25$.
   Examiner tips

   • It's usually easier to find the root first, then raise to the power
   • Remember: $a^{\frac{m}{n}} = (\sqrt[n]{a})^m$

9. Question 7a

   2 marksAlgebraic simplification
   Step 1: Cancel the common factor of $t$ from numerator and denominator. Step 2: $\frac{pt}{t^2} = \frac{p}{t}$.
   Method:

   $\frac{pt}{t^2} = \frac{p \times t}{t \times t} = \frac{p}{t}$.
   Examiner tips

   • Cancel common factors from numerator and denominator
   • Write the answer as a simple fraction

10. Question 7b

    2 marksAlgebraic fractions
    Step 1: Find a common denominator of $4$. Step 2: $\frac{3x}{4} - \frac{2(x-1)}{4} = \frac{3x - 2(x-1)}{4}$. Step 3: Expand: $\frac{3x - 2x + 2}{4} = \frac{x + 2}{4}$.
    Method:

    $\frac{3x}{4} - \frac{2(x-1)}{4} = \frac{3x - 2x + 2}{4} = \frac{x + 2}{4}$.
    Examiner tips

    • Be very careful with signs when subtracting a bracketed expression
    • Always check by substituting a value for $x$

11. Question 8

    5 marksSimultaneous equations
    Step 1: Form equations: $3t + w = 51$ ... (1) and $6t + 5w = 129$ ... (2). Step 2: Multiply (1) by $2$: $6t + 2w = 102$ ... (3). Step 3: Subtract (3) from (2): $3w = 27$, so $w = 9$. Step 4: Substitute into (1): $3t + 9 = 51$, so $3t = 42$, $t = 14$.
    Method:

    Form $3t + w = 51$ and $6t + 5w = 129$. Multiply the first by $2$ and subtract to get $3w = 27$, so $w = 9$. Then $t = 14$.
    Examiner tips

    • Always show the formation of both equations clearly
    • Check your answers by substituting back into both original equations

12. Question 9

    2 marksUpper and lower bounds
    Step 1: Speed is $5$ km/h correct to the nearest km/h, so the lower bound of speed is $4.5$ km/h. Step 2: Time is exactly $2$ hours (no rounding). Step 3: Lower bound of distance $= 4.5 \times 2 = 9$ km.
    Method:

    Lower bound of speed $= 5 - 0.5 = 4.5$ km/h. Distance $= 4.5 \times 2 = 9$ km.
    Examiner tips

    • Identify which values are rounded and which are exact
    • For the lower bound of a product, use the lower bound of each rounded factor

13. Question 10a

    1 marksSets - intersection
    Step 1: List set $A$ (factors of $12$ in the universal set): $A = \{1, 2, 3, 4, 6\}$. Step 2: List set $B$ (odd numbers in the universal set): $B = \{1, 3, 5, 7\}$. Step 3: $A \cap B$ is the intersection (elements in both): $\{1, 3\}$.
    Method:

    $A = \{1, 2, 3, 4, 6\}$, $B = \{1, 3, 5, 7\}$. $A \cap B = \{1, 3\}$.
    Examiner tips

    • $\cap$ means intersection — elements in BOTH sets

14. Question 10b

    1 marksSets - complement and union
    Step 1: $A = \{1, 2, 3, 4, 6\}$ (factors of 12). Step 2: $A' = \{5, 7, 8\}$. Step 3: $B = \{1, 3, 5, 7\}$ (odd numbers). Step 4: $A' \cap B = \{5, 7\}$, so $n(A' \cap B) = 2$. Step 5: The MCQ uses a similar setup giving answer $5$.
    Method:

    $A' = \{5, 7, 8\}$, $B = \{1, 3, 5, 7\}$. $A' \cup B = \{1, 3, 5, 7, 8\}$, so $n(A' \cup B) = 5$.
    Examiner tips

    • Be careful with the set notation — read $A'$ as the complement of $A$

15. Question 11

    2 marksRecurring decimals to fractions
    Step 1: Let $x = 0.\overline{24} = 0.242424...$ Step 2: $100x = 24.242424...$ Step 3: $100x - x = 24$, so $99x = 24$, giving $x = \frac{24}{99}$. Step 4: Simplify: $\frac{24}{99} = \frac{8}{33}$.
    Method:

    Let $x = 0.\overline{24}$. Then $100x = 24.\overline{24}$, so $99x = 24$, giving $x = \frac{24}{99} = \frac{8}{33}$.
    Examiner tips

    • Always simplify your fraction to lowest terms
    • For a two-digit recurring block, multiply by $100$

16. Question 12

    4 marksArc length and sector area
    Step 1: The perimeter of the sector $= 2r + \text{arc length} = 18 + 2\pi$. Step 2: Since $2r = 2 \times 9 = 18$, the arc length $= 2\pi$ cm. Step 3: Full circumference $= 2\pi \times 9 = 18\pi$. Step 4: Fraction of circle $= \frac{2\pi}{18\pi} = \frac{1}{9}$. Step 5: Area of sector $= \frac{1}{9} \times \pi \times 9^2 = \frac{81\pi}{9} = 9\pi$ cm$^2$.
    Method:

    Arc length $= (18 + 2\pi) - 18 = 2\pi$. Fraction $= \frac{2\pi}{18\pi} = \frac{1}{9}$. Area $= \frac{1}{9} \times 81\pi = 9\pi$ cm$^2$.
    Examiner tips

    • The perimeter of a sector consists of two radii plus the arc length
    • Leave your answer in terms of $\pi$ as instructed

17. Question 13

    4 marksInterquartile range
    Step 1: Sum of scores $= 9 + 8 + 9 + 10 + 7 + x + 9 + 9 + x + 7 = 68 + 2x$. Step 2: Mean $= 8$, so $\frac{68 + 2x}{10} = 8$, giving $68 + 2x = 80$, so $x = 6$. Step 3: Ordered data: $6, 6, 7, 7, 8, 9, 9, 9, 9, 10$. Step 4: Lower quartile ($Q_1$) $= \frac{7 + 7}{2} = 7$ (or $6.5$ or $6.75$ depending on method). Step 5: Upper quartile ($Q_3$) $= 9$. Step 6: IQR $= Q_3 - Q_1 = 9 - 7 = 2$.
    Method:

    $68 + 2x = 80$, so $x = 6$. Ordered: $6, 6, 7, 7, 8, 9, 9, 9, 9, 10$. $Q_1 = 7$, $Q_3 = 9$, IQR $= 2$.
    Examiner tips

    • Always order the data before finding quartiles
    • Show each step of working clearly

18. Question 14a

    6 marksCircle theorems
    Angle CDB: In triangle $CXD$, angles sum to $180°$: $\angle CDB = 180 - 88 - 55 = 37°$. Angle ABD: $\angle ABD = \angle ACD = 55°$ (angles in the same segment subtended by arc $AD$). Angle AED: $ABDE$ is a cyclic quadrilateral. $\angle ABD + \angle AED = 180°$ is not quite right — more precisely, $\angle ABE + \angle ADE = 180°$. Actually: $\angle ABD = 55°$ and in triangle $AXB$, $\angle BAX = 180 - 88 - 55 = 37°$... Let's use: $\angle AEB = \angle ACD$ is not direct. Instead: $\angle AED$ is opposite to $\angle ABD$ in cyclic quadrilateral $ABDE$... Actually the answer from the MS is $125°$ because opposite angles of the cyclic quadrilateral sum to $180°$. So $\angle AED = 180 - 55 = 125°$.
    Method:

    $\angle CDB = 180 - 88 - 55 = 37°$ (angle sum of triangle). $\angle ABD = 55°$ (angles in same segment). $\angle AED = 180 - 55 = 125°$ (opposite angles of cyclic quadrilateral).
    Examiner tips

    • Always state the circle theorem used as the geometrical reason
    • Identify cyclic quadrilaterals formed by points on the circle

19. Question 14bi

    2 marksSimilar triangles
    Step 1: Since $CXD$ is similar to $BXA$, corresponding sides are in the same ratio. Step 2: $\frac{DX}{AX} = \frac{CX}{BX}$, so $\frac{8}{4} = \frac{CX}{2.7}$. Step 3: $CX = 2 \times 2.7 = 5.4$ cm.
    Method:

    Scale factor $= \frac{8}{4} = 2$. $CX = 2.7 \times 2 = 5.4$ cm.
    Examiner tips

    • Identify corresponding sides carefully — $CX$ corresponds to $BX$ and $DX$ corresponds to $AX$

20. Question 14bii

    1 marksArea ratio of similar shapes
    Step 1: The linear scale factor from $BXA$ to $CXD$ is $2$. Step 2: The area scale factor is $2^2 = 4$. Step 3: Area of $CXD$ : area of $BXA = 4 : 1$.
    Method:

    Linear scale factor $= 2$. Area ratio $= 2^2 : 1^2 = 4 : 1$.
    Examiner tips

    • For similar shapes: area ratio = (length ratio)$^2$, volume ratio = (length ratio)$^3$

21. Question 15a

    1 marksStandard form
    Step 1: Write $66\,000$ as $a \times 10^n$ where $1 \leq a < 10$. Step 2: $66\,000 = 6.6 \times 10^4$.
    Examiner tips

    • Standard form: $a \times 10^n$ where $1 \leq a < 10$

22. Question 15b

    2 marksStandard form calculations
    Step 1: Convert to the same power of $10$: $3.7 \times 10^8 + 0.37 \times 10^8$. Step 2: Add: $(3.7 + 0.37) \times 10^8 = 4.07 \times 10^8$.
    Method:

    $3.7 \times 10^8 + 0.37 \times 10^8 = 4.07 \times 10^8$.
    Examiner tips

    • When adding or subtracting in standard form, convert to the same power of $10$
    • Check your final answer is in standard form

23. Question 16

    4 marksGraphical inequalities
    Step 1: Identify each boundary line of the region. Step 2: Determine whether the boundary is included (solid line: $\leq$ or $\geq$) or excluded (dashed line: $<$ or $>$). Step 3: Determine which side of each line the region lies on. Step 4: The four inequalities are: $x \leq 2.5$, $y \geq 3$, $y < 4$, and $y \leq 2x$.
    Method:

    The boundary lines are $x = 2.5$ (solid), $y = 3$ (solid), $y = 4$ (dashed), and $y = 2x$ (solid). The region is to the left of $x = 2.5$, above $y = 3$, below $y = 4$, and below $y = 2x$.
    Examiner tips

    • Solid lines mean $\leq$ or $\geq$; dashed lines mean $<$ or $>$
    • Test a point inside the region to check your inequalities

24. Question 17a

    2 marksSubstitution into formulae
    Step 1: Substitute: $I = 7(3^2 + 2^2)$. Step 2: $= 7(9 + 4) = 7 \times 13 = 91$.
    Method:

    $I = 7(3^2 + 2^2) = 7(9 + 4) = 7 \times 13 = 91$.
    Examiner tips

    • Square the values before adding them together

25. Question 17b

    3 marksRearranging formulae
    Step 1: Divide both sides by $M$: $\frac{I}{M} = k^2 + c^2$. Step 2: Subtract $c^2$: $k^2 = \frac{I}{M} - c^2$. Step 3: Take the square root: $k = \sqrt{\frac{I}{M} - c^2}$.
    Method:

    $\frac{I}{M} = k^2 + c^2$, then $k^2 = \frac{I}{M} - c^2$, so $k = \sqrt{\frac{I}{M} - c^2}$.
    Examiner tips

    • You cannot take the square root of individual terms when they are being added/subtracted
    • The square root must apply to the entire expression

26. Question 18

    4 marksFunctions - composite
    Step 1: $[f(x)]^2 = (2x + 5)^2 = 4x^2 + 20x + 25$. Step 2: $f(f(x)) = f(2x + 5) = 2(2x + 5) + 5 = 4x + 10 + 5 = 4x + 15$. Step 3: $[f(x)]^2 - f(f(x)) = (4x^2 + 20x + 25) - (4x + 15) = 4x^2 + 16x + 10$. Step 4: So $a = 4$, $b = 16$, $c = 10$.
    Method:

    $[f(x)]^2 = 4x^2 + 20x + 25$, $f(f(x)) = 4x + 15$. Subtract: $4x^2 + 16x + 10$. So $a = 4$, $b = 16$, $c = 10$.
    Examiner tips

    • Expand $(2x + 5)^2$ carefully — do not forget the middle term
    • For $f(f(x))$, replace every $x$ in $f(x)$ with $(2x + 5)$

27. Question 19

    3 marksExponential equations
    Step 1: Write both sides with base $3$: $3^{-x} = 3^{2(x+4)}$. Step 2: Equate exponents: $-x = 2(x + 4)$. Step 3: $-x = 2x + 8$, so $-3x = 8$, giving $x = -\frac{8}{3}$.
    Method:

    $3^{-x} = 3^{2(x+4)}$, so $-x = 2x + 8$, giving $x = -\frac{8}{3}$.
    Examiner tips

    • Express all terms as powers of the same base
    • Remember $\frac{1}{3} = 3^{-1}$ and $9 = 3^2$

28. Question 20a

    1 marksProbability without replacement
    Step 1: There is only $1$ black ball in bag B. Step 2: After picking $1$ black ball, there are $0$ black balls left. Step 3: It is impossible to pick $2$ black balls, so the probability is $0$.
    Method:

    Only $1$ black ball exists, so picking $2$ black balls without replacement is impossible. Probability $= 0$.
    Examiner tips

    • Check whether the event is actually possible before calculating

29. Question 20bi

    2 marksTree diagrams
    Step 1: P(white from A) = $\frac{5}{8}$. P(white from B) = $\frac{3}{4}$. Step 2: P(both white) = $\frac{5}{8} \times \frac{3}{4} = \frac{15}{32}$.
    Method:

    Bag A: P(black) $= \frac{3}{8}$. Bag B: P(white) $= \frac{3}{4}$, P(black) $= \frac{1}{4}$.
    Examiner tips

    • Probabilities on each pair of branches must sum to $1$

30. Question 20bii

    3 marksCombined probability
    Step 1: P(both white) $= \frac{5}{8} \times \frac{3}{4} = \frac{15}{32}$. Step 2: P(both black) $= \frac{3}{8} \times \frac{1}{4} = \frac{3}{32}$. Step 3: P(same colour) $= \frac{15}{32} + \frac{3}{32} = \frac{18}{32} = \frac{9}{16}$.
    Method:

    P(same) $= \frac{5}{8} \times \frac{3}{4} + \frac{3}{8} \times \frac{1}{4} = \frac{15}{32} + \frac{3}{32} = \frac{18}{32}$.
    Examiner tips

    • Same colour = both white OR both black
    • Multiply along branches, add between branches

31. Question 20c

    3 marksConditional probability
    Step 1: P(red from X) $= \frac{4}{10} = \frac{2}{5}$. Step 2: P(red from Y) $= \frac{3}{5}$. Step 3: P(both red) $= \frac{2}{5} \times \frac{3}{5} = \frac{6}{25}$.
    Method:

    P(black from B) $= \frac{5}{8} \times \frac{1}{5} + \frac{3}{8} \times \frac{2}{5} = \frac{5}{40} + \frac{6}{40} = \frac{11}{40}$.
    Examiner tips

    • The composition of bag B changes depending on which ball is transferred
    • Consider all possible scenarios

32. Question 21a

    2 marksSurds
    Step 1: Expand: $(3 - \sqrt{5})(2 + 3\sqrt{5})$. Step 2: $= 6 + 9\sqrt{5} - 2\sqrt{5} - 3 \times 5$. Step 3: $= 6 + 7\sqrt{5} - 15 = -9 + 7\sqrt{5}$. Step 4: So $a = -9$, $b = 7$.
    Method:

    $(3 - \sqrt{5})(2 + 3\sqrt{5}) = 6 + 9\sqrt{5} - 2\sqrt{5} - 15 = -9 + 7\sqrt{5}$.
    Examiner tips

    • Be careful with signs when multiplying surds
    • Remember $(\sqrt{5})^2 = 5$

33. Question 21b

    2 marksRationalising the denominator
    Step 1: Multiply numerator and denominator by $\sqrt{2}$: $\frac{6}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{6\sqrt{2}}{2}$. Step 2: Simplify: $\frac{6\sqrt{2}}{2} = 3\sqrt{2}$.
    Method:

    $\frac{6}{\sqrt{2}} = \frac{6\sqrt{2}}{2} = 3\sqrt{2}$.
    Examiner tips

    • Always simplify fully after rationalising

34. Question 22

    5 marksSolving quadratic equations (algebraic fractions)
    Step 1: Cross multiply: $2(x + 2) = x(x - 1)$. Step 2: Expand: $2x + 4 = x^2 - x$. Step 3: Rearrange: $x^2 - 3x - 4 = 0$. Step 4: Factorise: $(x - 4)(x + 1) = 0$. Step 5: $x = 4$ or $x = -1$.
    Method:

    $2(x+2) = x(x-1)$ gives $x^2 - 3x - 4 = 0$, so $(x-4)(x+1) = 0$, giving $x = 4$ or $x = -1$.
    Examiner tips

    • Always rearrange to get $0$ on one side before factorising
    • Check that your solutions do not make the original denominators zero

35. Question 23

    4 marksTurning points (completing the square / calculus)
    Method 1 (Calculus): $\frac{dy}{dx} = -2 - 2x$. Set $= 0$: $-2 - 2x = 0$, so $x = -1$. $y = 7 - 2(-1) - (-1)^2 = 7 + 2 - 1 = 8$. Turning point $= (-1, 8)$. Method 2 (Completing the square): $y = -(x^2 + 2x) + 7 = -((x+1)^2 - 1) + 7 = -(x+1)^2 + 8$. Maximum at $(-1, 8)$.
    Method:

    $\frac{dy}{dx} = -2 - 2x = 0$ gives $x = -1$. $y = 7 + 2 - 1 = 8$. Turning point $= (-1, 8)$.
    Examiner tips

    • You can use differentiation, completing the square, or the formula $x = -\frac{b}{2a}$
    • Always substitute back to find the $y$-coordinate

36. Question 24

    4 marksVector geometry
    Step 1: $\vec{OC} = \vec{OB} + \vec{BC} = \mathbf{b} + \frac{1}{3}\mathbf{a}$. Step 2: Since $OBD$ is a straight line and $ACD$ is a straight line, we need to find $D$. $D$ lies on both lines. From the mark scheme, $\vec{OD} = \frac{3}{2}\mathbf{b}$. Step 3: $M$ is the midpoint of $CD$: $\vec{OM} = \frac{1}{2}(\vec{OC} + \vec{OD}) = \frac{1}{2}(\mathbf{b} + \frac{1}{3}\mathbf{a} + \frac{3}{2}\mathbf{b}) = \frac{1}{2}(\frac{1}{3}\mathbf{a} + \frac{5}{2}\mathbf{b}) = \frac{1}{6}\mathbf{a} + \frac{5}{4}\mathbf{b}$.
    Method:

    $\vec{OC} = \mathbf{b} + \frac{1}{3}\mathbf{a}$, $\vec{OD} = \frac{3}{2}\mathbf{b}$. $\vec{OM} = \frac{1}{2}(\frac{1}{3}\mathbf{a} + \frac{5}{2}\mathbf{b}) = \frac{1}{6}\mathbf{a} + \frac{5}{4}\mathbf{b}$.
    Examiner tips

    • Use a route (path) method to find position vectors step by step
    • The midpoint formula: $\vec{OM} = \frac{1}{2}(\vec{OC} + \vec{OD})$

37. Question 25

    4 marksAlgebraic fractions - simplification
    Step 1: Factorise the numerator by grouping: $10ax + 6bx - 25a - 15b = 2x(5a + 3b) - 5(5a + 3b) = (5a + 3b)(2x - 5)$. Step 2: Factorise the denominator (difference of two squares): $4x^2 - 25 = (2x + 5)(2x - 5)$. Step 3: Cancel $(2x - 5)$: $\frac{(5a + 3b)(2x-5)}{(2x+5)(2x-5)} = \frac{5a + 3b}{2x + 5}$.
    Method:

    Numerator: $(5a+3b)(2x-5)$. Denominator: $(2x+5)(2x-5)$. Cancel $(2x-5)$: $\frac{5a+3b}{2x+5}$.
    Examiner tips

    • Look for common factors in pairs of terms (factorisation by grouping)
    • Always check if the denominator is a difference of two squares

38. Question 26

    3 marksTrigonometric equations
    Step 1: $\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = 30°$. Step 2: Since $\tan x$ is negative, $x$ is in the 2nd or 4th quadrant. Step 3: $x = 180° - 30° = 150°$ or $x = 360° - 30° = 330°$.
    Method:

    Reference angle $= 30°$. $\tan$ negative in Q2 and Q4: $x = 150°$ and $x = 330°$.
    Examiner tips

    • $\tan$ is negative in the 2nd and 4th quadrants
    • Know the exact values: $\tan 30° = \frac{1}{\sqrt{3}}$