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Admissions tests / TMUA / Paper 1 / Trigonometry

Test standard. 12 questions, 12 marks, about 45 minutes.

TMUA Paper 1: Trigonometry, set 2

Sine and cosine rules, exact values, graphs and transformations of trigonometric functions, identities and equations over a stated interval.

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  • Answer all questions. No calculator.
  • Each question has exactly one correct answer.
  1. 11 mark

    In triangle PQR, PQ = 6 cm, PR = 10 cm, and angle QPR = 120 degrees. Find the exact length of QR.

    1. A 196 cm
    2. B 14 cm
    3. C 2 sqrt(19) cm
    4. D 2 sqrt(34) cm
  2. 21 mark

    In triangle ABC, angle BAC = 30 degrees, angle ACB = 120 degrees, and BC = 4 cm. Find the exact length of AB.

    1. A 4 cm
    2. B 16 cm
    3. C (4 sqrt(3))/3 cm
    4. D 4 sqrt(3) cm
  3. 31 mark

    In triangle ABC, angle BAC = 45 degrees, AC = 8 sqrt(2) cm, and BC = 10 cm (BC is opposite the given angle). How many distinct triangles ABC satisfy these measurements?

    1. A 2
    2. B 1
    3. C 0
    4. D Infinitely many
  4. 41 mark

    OABC is a tetrahedron (a solid with four triangular faces) in which OA = OC = 8 cm and angle AOC = 120 degrees. Find the exact length of edge AC.

    1. A 192 cm
    2. B 8 sqrt(2) cm
    3. C 8 sqrt(3) cm
    4. D 8 cm
  5. 51 mark

    A sector-shaped tile has radius 12 cm, and the angle at its centre is pi/6 radians. Find the exact perimeter of the tile (its two straight edges plus the curved arc).

    1. A 2 pi cm
    2. B (2 pi + 24) cm
    3. C (pi + 24) cm
    4. D 24 pi cm
  6. 61 mark

    A sector of a circle has radius 4 cm and the angle at the centre is 2pi/3 radians. Find the exact length of the chord that joins the two ends of the arc.

    1. A 4 cm
    2. B 4 sqrt(2) cm
    3. C 4 sqrt(3) cm
    4. D 48 cm
  7. 71 mark

    Find the exact value of 4 sin(30) cos(60) + tan^2(60).

    1. A 6
    2. B 4/3
    3. C 1 + sqrt(3)
    4. D 4
  8. 81 mark

    Find the period of the graph of y = 3 tan(x/2), where x is measured in degrees.

    1. A 90 degrees
    2. B 180 degrees
    3. C 360 degrees
    4. D 720 degrees
  9. 91 mark

    Given that sin(theta) + cos(theta) = 7/5, find the exact value of 2 sin(theta) cos(theta).

    1. A 24/25
    2. B 49/25
    3. C -24/25
    4. D 7/5
  10. 101 mark

    Given that tan(theta) = 5/12 and 180 < theta < 270 (degrees), find the exact value of sin(theta).

    1. A 5/13
    2. B -5/13
    3. C -12/13
    4. D 12/13
  11. 111 mark

    Solve tan(x) = -1 for 0 <= x <= 360 (degrees). Which of the following gives the complete solution set?

    1. A {45, 225}
    2. B {45, 315}
    3. C {135, 315}
    4. D {135, 225}
  12. 121 mark

    Solve sin^2(x) = 3 cos^2(x) for 0 <= x <= 360 (degrees). How many solutions does the equation have in this interval?

    1. A 2
    2. B 3
    3. C 6
    4. D 4

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. The cosine rule for the side opposite the given angle is QR^2 = PQ^2 + PR^2 - 2(PQ)(PR) cos(QPR).
    2. Here PQ = 6, PR = 10 and angle QPR = 120 degrees, an exact standard value with cos(120) = -1/2.
    3. So QR^2 = 6^2 + 10^2 - 2(6)(10)(-1/2) = 36 + 100 + 60 = 196.
    4. Taking the square root, QR = sqrt(196) = 14. Therefore the answer is B: 14 cm.
    • Why not A: Correctly finds QR^2 = 196 using the cosine rule but forgets to take the square root, giving the squared length rather than QR itself.
    • Why not C: Uses cos(120) = 1/2 instead of the correct -1/2, missing the sign of cosine in the second quadrant; this turns the correction term positive and gives QR^2 = 136 - 60 = 76.
    • Why not D: Omits the -2(PQ)(PR) cos(QPR) term from the cosine rule altogether, effectively using Pythagoras's theorem (QR^2 = PQ^2 + PR^2) even though angle QPR is not a right angle.

  2. Question 2Answer: D

    1. In triangle ABC, angle A = 30 degrees, angle C = 120 degrees, and side BC = 4 cm is opposite angle A.
    2. The sine rule gives BC/sin(A) = AB/sin(C), where AB is opposite angle C, so AB = BC x sin(C)/sin(A).
    3. sin(120) equals sin(180 - 120) = sin(60) = sqrt(3)/2, by the symmetry of the sine graph about 90 degrees; sin(30) = 1/2, both exact standard values.
    4. So AB = 4 x (sqrt(3)/2) / (1/2) = 4 x sqrt(3) = 4 sqrt(3). Therefore the answer is D: 4 sqrt(3) cm.
    • Why not A: Takes sin(120) to be 1/2 (the value of sin(30), not sin(120)), which collapses side AB to equal side BC.
    • Why not B: Assumes sides scale in direct proportion to their opposite angles (AB = BC x (angle ACB / angle BAC) = 4 x (120/30) = 16), rather than using the sine rule's ratio of sides to sines of angles.
    • Why not C: Inverts the sine rule ratio, computing BC x sin(BAC)/sin(ACB) instead of BC x sin(ACB)/sin(BAC), giving 4 x (1/2)/(sqrt(3)/2) = 4/sqrt(3), which rationalises to (4 sqrt(3))/3.

  3. Question 3Answer: A

    1. Two distinct triangles ABC exist exactly when b sin(A) < a < b, given angle A, adjacent side b and opposite side a.
    2. Here b = AC = 8 sqrt(2) and A = 45 degrees, an exact standard angle, so b sin(A) = 8 sqrt(2) x (sqrt(2)/2) = 8.
    3. So two triangles exist precisely when 8 < a < 8 sqrt(2) (roughly 8 < a < 11.3). Here a = BC = 10, which lies strictly inside this range.
    4. Therefore there are exactly 2 distinct triangles ABC satisfying the given measurements, so the answer is A: 2.
    • Why not B: Notices that BC = 10 is less than AC = 8 sqrt(2), and concludes (incorrectly) that this alone guarantees exactly one triangle, without also checking whether BC exceeds AC sin(A).
    • Why not C: Mistakenly uses sin(45) = 1 (the value of sin(90), not sin(45)), which inflates AC sin(A) to 8 sqrt(2) - larger than BC = 10 - leading to the false conclusion that no triangle exists.
    • Why not D: Misreads 'the ambiguous case' as meaning the number of triangles cannot be determined, rather than correctly using it to mean up to two specific triangles can exist.

  4. Question 4Answer: C

    1. Triangle OAC is a face of the tetrahedron, with OA = OC = 8 cm and the included angle AOC = 120 degrees.
    2. The cosine rule gives AC^2 = OA^2 + OC^2 - 2(OA)(OC) cos(AOC).
    3. cos(120) = -1/2, an exact standard value, so AC^2 = 64 + 64 - 2(8)(8)(-1/2) = 128 + 64 = 192.
    4. AC = sqrt(192) = sqrt(64 x 3) = 8 sqrt(3). Therefore the answer is C: 8 sqrt(3) cm.
    • Why not A: Correctly computes AC^2 = 192 using the cosine rule but forgets to take the square root, leaving the squared length as the final answer.
    • Why not B: Omits the -2(OA)(OC) cos(AOC) term from the cosine rule entirely, effectively using Pythagoras's theorem as if angle AOC were a right angle, giving AC^2 = 8^2 + 8^2 = 128.
    • Why not D: Uses cos(120) = 1/2 instead of the correct -1/2, giving AC^2 = 64 + 64 - 128(1/2) = 64.

  5. Question 5Answer: B

    1. The perimeter of a sector is the arc length plus its two straight (radius) edges: Perimeter = r theta + 2r.
    2. Here r = 12 cm and theta = pi/6 radians, so the arc length is r theta = 12 x (pi/6) = 2 pi.
    3. The two straight edges each have length r = 12, contributing 2 x 12 = 24.
    4. So the perimeter is 2 pi + 24. Therefore the answer is B: (2 pi + 24) cm.
    • Why not A: Finds only the arc length (r theta = 12 x pi/6 = 2 pi) and forgets that the perimeter of a sector also includes its two straight radii.
    • Why not C: Mistakenly carries the 1/2 factor from the sector-area formula into the arc-length formula, using (1/2) r theta instead of r theta, which gives an arc of pi instead of 2 pi.
    • Why not D: Computes the circumference of the whole circle (2 pi r = 24 pi) rather than the arc length of the sector, ignoring that the sector spans only a pi/6 fraction of the full circle.

  6. Question 6Answer: C

    1. The chord AB and the two radii OA and OB form a triangle with OA = OB = 4 cm and included angle AOB = 2pi/3 radians (120 degrees).
    2. The cosine rule gives AB^2 = OA^2 + OB^2 - 2(OA)(OB) cos(AOB).
    3. cos(120 degrees) = -1/2, an exact standard value, so AB^2 = 16 + 16 - 2(4)(4)(-1/2) = 32 + 16 = 48.
    4. AB = sqrt(48) = sqrt(16 x 3) = 4 sqrt(3). Therefore the answer is C: 4 sqrt(3) cm.
    • Why not A: Uses cos(120 degrees) = 1/2 instead of the correct -1/2, giving AB^2 = 32 - 32(1/2) = 16, so AB = 4.
    • Why not B: Omits the -2(OA)(OB) cos(AOB) term from the cosine rule entirely, effectively using Pythagoras's theorem as though angle AOB were a right angle, giving AB^2 = 16 + 16 = 32, so AB = 4 sqrt(2).
    • Why not D: Correctly computes AB^2 = 48 using the cosine rule but forgets to take the square root, leaving the squared length as the final answer.

  7. Question 7Answer: D

    1. sin(30) = 1/2 and cos(60) = 1/2, both exact standard values, so 4 sin(30) cos(60) = 4 x (1/2) x (1/2) = 1.
    2. tan(60) = sqrt(3), so tan^2(60) = 3.
    3. Adding the two parts: 1 + 3 = 4.
    4. Therefore the answer is D: 4.
    • Why not A: Swaps the angles in the first term, computing sin(60) and cos(30) in place of sin(30) and cos(60); since sin(60) cos(30) = 3/4 rather than sin(30) cos(60) = 1/4, this inflates the first term to 3, giving 3 + 3 = 6.
    • Why not B: Uses tan(30) in place of tan(60) - the wrong one of the two paired standard angles - so tan^2(30) = 1/3 is added instead of tan^2(60) = 3.
    • Why not C: Correctly evaluates the first term as 1 but forgets to square tan(60), adding tan(60) = sqrt(3) itself instead of tan^2(60) = 3.

  8. Question 8Answer: C

    1. The graph of y = tan(x) has period 180 degrees.
    2. For y = tan(kx), the period is 180/k degrees; here the argument is x/2, so k = 1/2.
    3. Period = 180 / (1/2) = 360 degrees.
    4. The factor of 3 outside the tangent only scales the output values (a vertical stretch) and does not affect the period. Therefore the answer is C: 360 degrees.
    • Why not A: Multiplies the standard tan period (180 degrees) by the coefficient 1/2 instead of dividing by it, treating the transformation as a compression rather than a stretch.
    • Why not B: Ignores the effect of the coefficient 1/2 inside tan(x/2) altogether and simply states the untransformed period of tan(x), 180 degrees.
    • Why not D: Uses the period rule 360/k (which applies to sin(kx) and cos(kx), whose base period is 360 degrees) instead of the correct rule 180/k for tan(kx), whose base period is 180 degrees.

  9. Question 9Answer: A

    1. Squaring both sides of sin(theta) + cos(theta) = 7/5 gives sin^2(theta) + 2 sin(theta) cos(theta) + cos^2(theta) = 49/25.
    2. Using the identity sin^2(theta) + cos^2(theta) = 1, this becomes 1 + 2 sin(theta) cos(theta) = 49/25.
    3. So 2 sin(theta) cos(theta) = 49/25 - 1 = 49/25 - 25/25 = 24/25.
    4. Therefore the answer is A: 24/25.
    • Why not B: Squares the given equation to get 49/25 but forgets that (sin theta + cos theta)^2 also contains the sin^2(theta) + cos^2(theta) = 1 term, so never subtracts it off.
    • Why not C: Rearranges the identity with a sign slip, computing 1 - 49/25 instead of 49/25 - 1, giving the negative of the correct value.
    • Why not D: Assumes 2 sin(theta) cos(theta) can be read directly from sin(theta) + cos(theta) without squaring the given equation at all.

  10. Question 10Answer: B

    1. tan(theta) = 5/12, so consider a right-angled triangle with opposite side 5 and adjacent side 12; by Pythagoras, the hypotenuse is sqrt(5^2 + 12^2) = sqrt(169) = 13.
    2. This gives the magnitudes sin(theta) = 5/13 and cos(theta) = 12/13 (check: tan = sin/cos = (5/13)/(12/13) = 5/12, as required).
    3. Since 180 < theta < 270, theta is in the third quadrant, where both sine and cosine are negative.
    4. So sin(theta) = -5/13. Therefore the answer is B: -5/13.
    • Why not A: Correctly builds the right-angled triangle with opposite 5, adjacent 12 and hypotenuse 13 (5-12-13), but forgets that sine is negative for theta in the third quadrant.
    • Why not C: Swaps which side is opposite and which is adjacent, using the '12' side for sine instead of the '5' side, while correctly keeping the negative sign for the third quadrant.
    • Why not D: Makes both errors at once: swaps sine and cosine's sides (using 12 instead of 5) and forgets the negative sign required in the third quadrant.

  11. Question 11Answer: C

    1. tan(x) = -1 requires the reference angle to be 45 degrees (since tan(45) = 1, an exact standard value), with tan(x) negative.
    2. tan(x) is negative in the second and fourth quadrants.
    3. In the second quadrant: x = 180 - 45 = 135. In the fourth quadrant: x = 360 - 45 = 315.
    4. Both lie in 0 <= x <= 360, so the complete solution set is {135, 315}, answer C.
    • Why not A: Solves tan(x) = 1 instead of tan(x) = -1 (dropping the negative sign), giving the first- and third-quadrant solutions 45 and 225, where tangent is positive rather than negative.
    • Why not B: Correctly finds the fourth-quadrant solution 315 but pairs it with 45 (first quadrant, where tangent is positive), instead of the matching second-quadrant solution 135.
    • Why not D: Correctly finds the second-quadrant solution 135 but pairs it with 225 (third quadrant, where tangent is positive, not negative), instead of the matching fourth-quadrant solution 315.

  12. Question 12Answer: D

    1. If cos(x) = 0, then sin^2(x) = 1 (from sin^2(x) + cos^2(x) = 1), but the equation would need 1 = 3(0) = 0, which is false, so cos(x) is never zero for a solution and it is valid to divide through by cos^2(x).
    2. Dividing gives tan^2(x) = 3, so tan(x) = sqrt(3) or tan(x) = -sqrt(3).
    3. tan(x) = sqrt(3) (reference angle 60, exact standard value): x = 60 or x = 240 (quadrants 1 and 3, where tangent is positive).
    4. tan(x) = -sqrt(3): x = 120 or x = 300 (quadrants 2 and 4, where tangent is negative).
    5. This gives four solutions in total: 60, 120, 240 and 300. Therefore the answer is D: 4.
    • Why not A: Rearranges to tan^2(x) = 3 correctly, but only takes tan(x) = sqrt(3) (the positive square root), missing the equally valid tan(x) = -sqrt(3) branch and its two solutions.
    • Why not B: Reaches tan(x) = sqrt(3) or tan(x) = -sqrt(3) but loses one solution while listing them, for instance quoting only one of the two solutions from the negative branch.
    • Why not C: Wrongly includes x = 90 and x = 270 (where cos(x) = 0) as extra solutions, without checking that dividing by cos(x) requires cos(x) != 0; at these values sin^2(x) = 1 while 3cos^2(x) = 0, so the original equation is not actually satisfied there.

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