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Admissions tests / ESAT / Maths 2 / Coordinate geometry and trigonometry

Stretch. 15 questions, 15 marks, about 30 minutes.

ESAT Mathematics 2: Coordinate geometry and trigonometry, set 4

Straight lines, circles, tangents and normals, intersections, the sine and cosine rules, exact values, trigonometric graphs, identities and equations.

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  • Answer all questions. No calculator.
  • Each question has exactly one correct answer.
  • This is a stretch set: several questions combine two specification points, and the fastest route is rarely the first one you see.
  1. 11 mark

    Find the values of k for which the line y = 2x + k is a tangent to the circle x^2 + y^2 = 5.

    1. A k = 2sqrt(5) or k = -2sqrt(5)
    2. B k = 5 or k = -5
    3. C k = sqrt(5) or k = -sqrt(5)
    4. D k = 5sqrt(5) or k = -5sqrt(5)
  2. 21 mark

    Points A and B lie on a circle with centre O, and angle AOB = 100 degrees. Points P and Q both lie on the major arc AB (distinct from each other and from A and B). Find angle AQB.

    1. A 100 degrees
    2. B 25 degrees
    3. C Cannot be determined without knowing exactly where Q lies
    4. D 50 degrees
  3. 31 mark

    In triangle ABC, angle A = 30 degrees, side a (opposite A) = 5 cm, and side b (opposite B) = 5sqrt(3) cm. Use the sine rule to find the possible value(s) of angle B.

    1. A B = 60 degrees or B = 120 degrees
    2. B B = 60 degrees only
    3. C B = 120 degrees only
    4. D B = 90 degrees
  4. 41 mark

    A sector of a circle has radius 6 cm and angle pi/3 radians at the centre. Find the area of the corresponding minor segment, giving your answer in the form a*pi - b*sqrt(3), in cm^2.

    1. A 24pi - 36sqrt(3) cm^2
    2. B 6pi cm^2
    3. C 6pi - 9sqrt(3) cm^2
    4. D 6pi - 9 cm^2
  5. 51 mark

    The circle C passes through the points A(1, 1), B(5, 1) and D(1, 5). Find the coordinates of the centre of C.

    1. A (7/3, 7/3)
    2. B (3, 1)
    3. C (1, 3)
    4. D (3, 3)
  6. 61 mark

    Solve 2 sin^2(x) = 1 + cos(x) for 0 <= x <= 360 degrees, giving all solutions.

    1. A x = 60 or x = 300 degrees
    2. B x = 60, 180 or 300 degrees
    3. C x = 180 degrees only
    4. D x = 60, 120 or 180 degrees
  7. 71 mark

    Two straight roads meet at a junction J at an angle of 120 degrees. A cyclist rides 4 km from J along one road, and a walker walks 3 km from J along the other road. Find the direct distance between the cyclist and the walker, in exact surd form.

    1. A sqrt(31) km
    2. B 5 km
    3. C sqrt(37) km
    4. D sqrt(13) km
  8. 81 mark

    A circle has equation x^2 + y^2 - 4x + 6y - 12 = 0. The chord AB of the circle lies on the line x = 1. Find the length of AB.

    1. A 4sqrt(6)
    2. B 2sqrt(6)
    3. C 8
    4. D 2sqrt(26)
  9. 91 mark

    Solve cos(3x) = 1 for 0 <= x <= 360 degrees, giving all solutions.

    1. A x = 0, 120 or 240 degrees
    2. B x = 0 or 120 degrees
    3. C x = 0 or 360 degrees
    4. D x = 0, 120, 240 or 360 degrees
  10. 101 mark

    Line l passes through the centre of the circle x^2 + y^2 - 8x + 2y + 8 = 0 and is parallel to the line 3x - 4y = 7. Find the equation of l, in the form ax + by = c.

    1. A 4x + 3y = 13
    2. B 3x - 4y = 16
    3. C 3x - 4y = -16
    4. D 3x - 4y = 20
  11. 111 mark

    AB is a diameter of a circle with centre O. C is a point on the circle, and the tangent to the circle at B meets line AC extended at the point D. Given that angle BAC = 35 degrees, find angle ADB.

    1. A 55 degrees
    2. B 90 degrees
    3. C 35 degrees
    4. D 20 degrees
  12. 121 mark

    A sector of a circle has radius 10 cm and angle pi/3 radians at the centre. Find the length of the chord joining the two ends of the arc.

    1. A 10sqrt(2) cm
    2. B 10pi/3 cm
    3. C 10 cm
    4. D 5sqrt(6) cm
  13. 131 mark

    Solve 2 sin(x) cos(x) = cos(x) for 0 <= x <= 360 degrees, giving all solutions.

    1. A x = 30, 90, 210 or 270 degrees
    2. B x = 30, 90, 150 or 270 degrees
    3. C x = 30 or 150 degrees
    4. D x = 0, 30, 150 or 180 degrees
  14. 141 mark

    The line y = 2x + 1 intersects the circle x^2 + y^2 = 20 at two points P and Q. Find the coordinates of the midpoint of PQ.

    1. A (2/5, 9/5)
    2. B (-4/5, -3/5)
    3. C (0, 1)
    4. D (-2/5, 1/5)
  15. 151 mark

    In triangle ABC, angle A = 60 degrees, side a (opposite A) = 2sqrt(3) cm, and side c (opposite C) = 2 cm. Use the sine rule to find angle C, rejecting any value that does not give a valid triangle.

    1. A 150 degrees
    2. B 30 degrees or 150 degrees
    3. C 30 degrees
    4. D 60 degrees

Worked solutions

Every question below carries the reasoning, not just the answer. The official material for this test publishes a correct option letter and nothing else.

  1. Question 1Answer: B

    1. Rewrite the line as 2x - y + k = 0. The circle x^2 + y^2 = 5 has centre (0, 0) and radius sqrt(5).
    2. The line is a tangent exactly when the perpendicular distance from the centre to the line equals the radius: |2(0) - (0) + k| / sqrt(2^2 + (-1)^2) = sqrt(5).
    3. This gives |k| / sqrt(5) = sqrt(5), so |k| = 5, i.e. k = 5 or k = -5.
    4. (Check: substituting y = 2x + 5 into x^2 + y^2 = 5 gives 5x^2 + 20x + 20 = 0, i.e. 5(x + 2)^2 = 0, a repeated root, confirming tangency.) So the answer is B.
    • Why not A: Uses the line's gradient, 2, in place of sqrt(1 + gradient^2) in the perpendicular distance formula, setting |k|/2 = sqrt(5) instead of |k|/sqrt(5) = sqrt(5), giving k = 2sqrt(5) or -2sqrt(5).
    • Why not C: Substitutes y = 2x + k into the circle's equation but expands (2x + k)^2 as 4x^2 + k^2, dropping the cross term 4kx; setting the discriminant of the resulting (incorrect) quadratic 5x^2 + (k^2 - 5) = 0 to zero then gives k = sqrt(5) or -sqrt(5) instead of 5 or -5.
    • Why not D: Sets the perpendicular distance from the centre equal to r^2 = 5 rather than to the actual radius r = sqrt(5), solving |k|/sqrt(5) = 5 to get k = 5sqrt(5) or -5sqrt(5).

  2. Question 2Answer: D

    1. The angle at the centre of a circle is twice the angle at the circumference, for angles subtended by the same arc, with the circumference point on the major arc: so angle APB = (1/2)(100) = 50 degrees for any point P on the major arc.
    2. Angles in the same segment subtended by the same arc are equal, so any other point Q on the same major arc also gives angle AQB = angle APB.
    3. So angle AQB = 50 degrees regardless of exactly where Q lies on the major arc, option D.
    • Why not A: Forgets that the angle at the centre is twice the angle at the circumference subtending the same arc, and simply carries the given centre angle of 100 degrees over to angle AQB unchanged.
    • Why not B: Correctly halves the centre angle once to find the circumference angle at any one point (50 degrees), but then mistakenly halves again on the assumption that a second point Q must give half of the first point's angle, rather than the same angle.
    • Why not C: Misunderstands the circle theorem that angles subtended by the same arc, from points in the same segment, are all equal; assumes the angle depends on exactly where Q sits on the major arc rather than being fixed for every such point.

  3. Question 3Answer: A

    1. By the sine rule, a/sin A = b/sin B, so sin B = b sin A / a = (5sqrt(3))(sin 30) / 5 = (5sqrt(3))(1/2)/5 = sqrt(3)/2.
    2. Since sin B = sqrt(3)/2 = sin 60 degrees, and sine is also positive in the second quadrant, B = 60 degrees or B = 180 - 60 = 120 degrees.
    3. Check both are valid triangles: if B = 60, C = 180 - 30 - 60 = 90 degrees (positive, valid). If B = 120, C = 180 - 30 - 120 = 30 degrees (positive, valid). Both give a genuine triangle, so both are correct, option A.
    • Why not B: Finds sin B = sqrt(3)/2 correctly but only takes the acute solution B = 60 degrees, forgetting that the sine rule can give a second, obtuse solution for a non-included angle.
    • Why not C: Assumes the obtuse solution must be the only valid one, perhaps reasoning that since side b (opposite B) is longer than side a (opposite A), angle B must be the larger, obtuse angle, and misses that the acute solution B = 60 also produces a valid triangle.
    • Why not D: Computes sin B = sqrt(3)/2 correctly, giving B = 60 or 120, but then reports the resulting third angle for the B = 60 case, C = 180 - 30 - 60 = 90 degrees, mistaking it for the value of B that was asked for.

  4. Question 4Answer: C

    1. The sector area, with angle theta in radians, is (1/2) r^2 theta = (1/2)(6^2)(pi/3) = (1/2)(36)(pi/3) = 6pi cm^2.
    2. The triangle formed by the two radii and the chord has area (1/2) r^2 sin(theta) = (1/2)(36) sin(60 degrees) = 18 x (sqrt(3)/2) = 9sqrt(3) cm^2, since pi/3 radians = 60 degrees.
    3. The minor segment is the sector minus this triangle: 6pi - 9sqrt(3) cm^2, option C.
    • Why not A: Uses the diameter (12 cm) in place of the radius in both the sector-area and triangle-area formulas, giving a sector area of 24pi and a triangle area of 36sqrt(3), instead of using the radius of 6 cm throughout.
    • Why not B: Finds the correct sector area, 6pi cm^2, but forgets that the segment is the sector area minus the triangle area, and gives the sector area alone as the final answer.
    • Why not D: Confuses sin(60 degrees) = sqrt(3)/2 with cos(60 degrees) = 1/2 when finding the triangle's area, computing (1/2)(36)(1/2) = 9 instead of (1/2)(36)(sqrt(3)/2) = 9sqrt(3).

  5. Question 5Answer: D

    1. The perpendicular from the centre to any chord bisects that chord, so the centre lies on the perpendicular bisector of every chord of the circle.
    2. Chord AB joins (1,1) and (5,1), which share the same y-coordinate, so AB is horizontal; its perpendicular bisector is the vertical line through its midpoint (3, 1), i.e. x = 3.
    3. Chord AD joins (1,1) and (1,5), which share the same x-coordinate, so AD is vertical; its perpendicular bisector is the horizontal line through its midpoint (1, 3), i.e. y = 3.
    4. The centre lies on both perpendicular bisectors, so it is their intersection: (3, 3), option D.
    • Why not A: Averages the three given points' coordinates as if the centre were their centroid, ((1+5+1)/3, (1+1+5)/3) = (7/3, 7/3), rather than finding the point equidistant from all three, which requires intersecting perpendicular bisectors.
    • Why not B: Finds the midpoint of chord AB, (3, 1), and stops there, forgetting that the midpoint of a chord only pins down the centre's position along the perpendicular bisector of that one chord; a second chord is needed to fix the centre exactly.
    • Why not C: Finds the midpoint of chord AD, (1, 3), and stops there, making the same error as only using one chord's perpendicular bisector rather than intersecting it with a second.

  6. Question 6Answer: B

    1. Use sin^2(x) = 1 - cos^2(x) to rewrite the equation entirely in terms of cos(x): 2(1 - cos^2(x)) = 1 + cos(x).
    2. Expand and rearrange: 2 - 2cos^2(x) = 1 + cos(x), so 2cos^2(x) + cos(x) - 1 = 0.
    3. Factorise: (2cos(x) - 1)(cos(x) + 1) = 0, so cos(x) = 1/2 or cos(x) = -1.
    4. cos(x) = 1/2 gives x = 60 or x = 360 - 60 = 300 degrees (cosine positive in the first and fourth quadrants). cos(x) = -1 gives x = 180 degrees. So the full solution set is x = 60, 180, 300 degrees, option B.
    • Why not A: Correctly factors to (2cos x - 1)(cos x + 1) = 0 but only solves the first bracket for cos x = 1/2, treating the second bracket cos x = -1 as if it gave no solution in range, instead of the valid x = 180 degrees.
    • Why not C: Uses the identity sin^2(x) = cos^2(x) - 1 (a sign error; the correct identity is sin^2(x) = 1 - cos^2(x)), which turns the equation into 2cos^2(x) - cos(x) - 3 = 0; factoring gives cos(x) = 3/2 (impossible) or cos(x) = -1, leaving only x = 180 degrees.
    • Why not D: Correctly finds cos(x) = 1/2 as one solution branch, but uses the quadrant rule for a positive SINE value (first and second quadrants) rather than for a positive COSINE value (first and fourth quadrants), giving x = 60 or 120 degrees instead of x = 60 or 300 degrees.

  7. Question 7Answer: C

    1. Let d be the distance between the cyclist and the walker. By the cosine rule, d^2 = 4^2 + 3^2 - 2(4)(3) cos(120 degrees).
    2. cos(120 degrees) = -cos(60 degrees) = -1/2, since 120 degrees is in the second quadrant where cosine is negative.
    3. d^2 = 16 + 9 - 24(-1/2) = 25 + 12 = 37.
    4. So d = sqrt(37) km, option C.
    • Why not A: Omits the factor of 2 in the cosine rule, computing d^2 = 4^2 + 3^2 - (4)(3)cos(120) = 25 - 12(-1/2) = 25 + 6 = 31 instead of using 2(4)(3) = 24 in the subtracted term.
    • Why not B: Applies Pythagoras' theorem directly, d^2 = 4^2 + 3^2, as though the angle at J were 90 degrees, ignoring that the actual angle between the roads is 120 degrees.
    • Why not D: Uses cos(120 degrees) = 1/2 instead of the correct negative value -1/2 (forgetting that cosine is negative for obtuse angles), computing d^2 = 25 - 24(1/2) = 25 - 12 = 13.

  8. Question 8Answer: A

    1. Complete the square: x^2 - 4x = (x - 2)^2 - 4, and y^2 + 6y = (y + 3)^2 - 9. So (x-2)^2 - 4 + (y+3)^2 - 9 - 12 = 0, giving (x-2)^2 + (y+3)^2 = 25. The circle has centre (2, -3) and radius 5.
    2. The perpendicular from the centre to the chord x = 1 has length equal to the horizontal distance from x = 2 to x = 1, which is 1.
    3. The perpendicular from the centre to a chord bisects it, so half of AB has length sqrt(radius^2 - distance^2) = sqrt(25 - 1) = sqrt(24) = 2sqrt(6).
    4. So the full chord AB = 2 x 2sqrt(6) = 4sqrt(6), option A.
    • Why not B: Correctly finds the radius (5) and the perpendicular distance from the centre to the chord (1), computing the half-chord length sqrt(25 - 1) = sqrt(24) = 2sqrt(6), but forgets that this is only half of AB and does not double it.
    • Why not C: Uses the centre's y-coordinate, -3, as the perpendicular distance from the centre to the line x = 1, instead of the correct horizontal distance from the centre's x-coordinate (2) to the line (1), which is 1; this gives sqrt(25 - 9) x 2 = 4 x 2 = 8.
    • Why not D: Adds the square of the perpendicular distance to the square of the radius instead of subtracting it, computing 2sqrt(25 + 1) = 2sqrt(26) instead of 2sqrt(25 - 1).

  9. Question 9Answer: D

    1. cos(theta) = 1 exactly when theta = 0, 360, 720, 1080, ... (every multiple of 360 degrees).
    2. Here theta = 3x, and since 0 <= x <= 360, the range for theta is 0 <= 3x <= 1080.
    3. The multiples of 360 within this range are 0, 360, 720 and 1080, so 3x = 0, 360, 720, 1080.
    4. Dividing by 3: x = 0, 120, 240, 360 degrees, option D.
    • Why not A: Correctly finds that 3x = 0, 360, 720 all satisfy cos(3x) = 1 within the scaled range, giving x = 0, 120, 240, but overlooks that 3x = 1080 (giving x = 360) is also a valid solution, since 3 x 360 = 1080 is included at the closed upper endpoint of the range for 3x.
    • Why not B: Solves cos(3x) = 1 using 0 <= 3x <= 360 (the same bounds as the original interval for x), instead of correctly scaling the bounds to 0 <= 3x <= 1080 to match 0 <= x <= 360; this loses two of the four solutions.
    • Why not C: Assumes the substitution x -> 3x does not change how many times the graph repeats within the given range, and simply carries over the two solutions of cos(x) = 1 in 0 <= x <= 360 (namely x = 0 and x = 360) unchanged, without accounting for the extra repetitions caused by the argument 3x.

  10. Question 10Answer: B

    1. Complete the square on the circle's equation: x^2 - 8x = (x-4)^2 - 16, and y^2 + 2y = (y+1)^2 - 1. So (x-4)^2 - 16 + (y+1)^2 - 1 + 8 = 0, giving (x-4)^2 + (y+1)^2 = 9, with centre (4, -1).
    2. Rearranging 3x - 4y = 7 gives y = (3/4)x - 7/4, so this line has gradient 3/4. A line parallel to it also has gradient 3/4.
    3. Using the centre (4, -1): y - (-1) = (3/4)(x - 4), so y + 1 = (3/4)x - 3, giving y = (3/4)x - 4.
    4. Multiply by 4 and rearrange: 4y = 3x - 16, so 3x - 4y = 16, option B.
    • Why not A: Correctly finds the centre (4, -1) but uses the negative reciprocal of 3/4 (i.e. -4/3) as the gradient of l, as though l needed to be perpendicular to the given line rather than parallel to it.
    • Why not C: Finds the centre of the circle using centre = (c/2, d/2) directly from the general form x^2+y^2+cx+dy+e=0, instead of the correct centre = (-c/2, -d/2); with c = -8 and d = 2, this gives (-4, 1) instead of the correct (4, -1), and carrying this wrong centre through the rest of the working gives 3x - 4y = -16.
    • Why not D: Correctly finds the centre (4, -1) and the correct gradient 3/4, but makes an expansion error, writing (3/4)(x - 4) as (3/4)x - 4 instead of (3/4)x - 3 (forgetting to multiply the 4 by 3/4), which shifts the final constant and gives 3x - 4y = 20 instead of 3x - 4y = 16.

  11. Question 11Answer: A

    1. Since AB is a diameter, the angle in a semicircle theorem gives angle ACB = 90 degrees.
    2. The tangent to a circle is perpendicular to the radius at the point of contact; since O lies on AB, the tangent at B is perpendicular to AB, so angle ABD = 90 degrees, where D lies on ray AC extended.
    3. In triangle ABD, angle DAB = angle CAB = 35 degrees (the same angle, since D lies on line AC extended) and angle ABD = 90 degrees.
    4. So angle ADB = 180 - 90 - 35 = 55 degrees, option A.
    • Why not B: Assumes the tangent creates the right angle directly at D (angle ADB = 90 degrees) rather than at B (angle ABD = 90 degrees, from the tangent being perpendicular to the diameter AB), so never applies the angle sum of triangle ABD.
    • Why not C: Misapplies the alternate segment theorem, assuming it hands over angle ADB directly as equal to the given angle BAC = 35 degrees, when the theorem actually relates the tangent-chord angle to a different angle (angle DBC), not to angle ADB.
    • Why not D: Correctly identifies that the angle 35 degrees appears in the reasoning, but subtracts it twice from the right angle (90 - 35 - 35 = 20) instead of using the angle sum of triangle ABD (180 - 90 - 35 = 55), effectively confusing two different triangles' angle totals.

  12. Question 12Answer: C

    1. The angle pi/3 radians equals 60 degrees. The two radii to the ends of the arc, each of length 10 cm, together with the chord, form an isosceles triangle with the angle between the equal sides equal to 60 degrees.
    2. An isosceles triangle with its included angle equal to 60 degrees is in fact equilateral (since the two base angles must also be equal to each other and to (180-60)/2 = 60 degrees), so the chord has the same length as the two radii, 10 cm.
    3. (Check with the cosine rule: chord^2 = 10^2 + 10^2 - 2(10)(10)cos(60) = 200 - 200(1/2) = 100, so chord = 10, confirming the shortcut.) So the answer is C.
    • Why not A: Misconverts pi/3 radians, treating it as a right angle (90 degrees) rather than 60 degrees, and then uses Pythagoras on the resulting (incorrect) right-angled isosceles triangle with two sides of 10 cm, giving a chord of 10sqrt(2) cm.
    • Why not B: Confuses the straight-line chord length with the curved arc length, computing r x theta = 10 x (pi/3) = 10pi/3 instead of finding the length of the straight segment joining the two ends of the arc.
    • Why not D: Applies the cosine rule but omits the factor of 2 in the term 2ab cos(C), computing chord^2 = 10^2 + 10^2 - (10)(10)cos(60 degrees) = 200 - 50 = 150 instead of 200 - 100(1/2) = 100, giving 5sqrt(6) instead of 10.

  13. Question 13Answer: B

    1. Rearrange as 2sin(x)cos(x) - cos(x) = 0, and factorise: cos(x)(2sin(x) - 1) = 0.
    2. So either cos(x) = 0, giving x = 90 or x = 270 degrees, or sin(x) = 1/2, giving x = 30 or x = 180 - 30 = 150 degrees.
    3. Combining both branches, and keeping every root of the factorised equation (rather than dividing by cos(x), which would silently discard the cos(x) = 0 solutions), the full solution set is x = 30, 90, 150, 270 degrees, option B.
    • Why not A: Correctly solves cos(x) = 0 to get x = 90, 270 degrees, but for sin(x) = 1/2 uses the wrong second-quadrant reference, computing 180 + 30 = 210 instead of 180 - 30 = 150 degrees, forgetting that sine (not cosine) is positive throughout the second quadrant.
    • Why not C: Divides both sides of the original equation by cos(x), reducing it to 2sin(x) = 1, and so finds only x = 30 and x = 150 degrees; this loses the solutions where cos(x) = 0 (x = 90, 270), which are lost whenever an equation is divided through by an expression that can itself equal zero.
    • Why not D: Confuses the zeros of cos(x) with the zeros of sin(x), incorrectly solving cos(x) = 0 as x = 0, 180 degrees instead of the correct x = 90, 270 degrees.

  14. Question 14Answer: D

    1. Substitute y = 2x + 1 into x^2 + y^2 = 20: x^2 + (2x+1)^2 = 20, so x^2 + 4x^2 + 4x + 1 = 20, giving 5x^2 + 4x - 19 = 0.
    2. If the two roots of this quadratic are x1 and x2 (the x-coordinates of P and Q), then by Vieta's formulas x1 + x2 = -4/5, without needing to solve for either root individually.
    3. The midpoint's x-coordinate is the average (x1+x2)/2 = (-4/5)/2 = -2/5.
    4. Substitute into the line to find the y-coordinate: y = 2(-2/5) + 1 = -4/5 + 5/5 = 1/5. So the midpoint is (-2/5, 1/5), option D.
    • Why not A: Uses the sum of the quadratic's roots as +4/5 rather than -4/5 (a sign slip in -b/a, using b/a instead), giving x-coordinate of the midpoint (4/5)/2 = 2/5 and, via the line, y = 2(2/5) + 1 = 9/5.
    • Why not B: Correctly finds that the roots of 5x^2 + 4x - 19 = 0 sum to -4/5, but forgets that the midpoint's x-coordinate is HALF this sum (since it is the average of the two roots), and uses -4/5 directly as the x-coordinate instead of -2/5.
    • Why not C: Expands (2x+1)^2 as 4x^2 + 1, dropping the middle term 4x, which turns the substituted equation into 5x^2 - 19 = 0; this (incorrect) quadratic has no x-term, so its roots sum to zero, giving a midpoint x-coordinate of 0 and, via the line, y = 1.

  15. Question 15Answer: C

    1. By the sine rule, a/sin A = c/sin C, so sin C = c sin A / a = (2)(sin 60) / (2sqrt(3)) = (2)(sqrt(3)/2) / (2sqrt(3)) = sqrt(3) / (2sqrt(3)) = 1/2.
    2. sin C = 1/2 gives C = 30 degrees or C = 180 - 30 = 150 degrees.
    3. Check each against the triangle's angle sum with angle A = 60 degrees: if C = 150, then A + C = 210 degrees, already more than 180 degrees, which is impossible; if C = 30, then A + C = 90 degrees, leaving angle B = 90 degrees, a valid triangle.
    4. So the only valid solution is C = 30 degrees, option C.
    • Why not A: Finds sin C = 1/2 correctly but takes only the obtuse solution, C = 150 degrees, without checking that angle A + angle C = 60 + 150 = 210 degrees already exceeds 180 degrees, which makes this solution impossible.
    • Why not B: Correctly finds both candidate values, C = 30 or C = 150 degrees, from sin C = 1/2, but does not check either against the triangle's angle sum, so fails to reject the impossible obtuse case.
    • Why not D: Finds sin C = 1/2 correctly, but reads off C = 60 degrees, confusing the identity sin(30) = 1/2 with cos(60) = 1/2, and mistakenly reports the angle associated with 1/2 in the cosine table rather than the sine table.

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