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

Test standard. 15 questions, 15 marks, about 23 minutes.

ESAT Mathematics 2: Coordinate geometry and trigonometry, set 2

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

Download the questions (PDF) Download with worked solutions (PDF)

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

    Line L1 passes through the points (0, 4) and (3, 10). Line L2 has equation 4x - 2y + 7 = 0. State the relationship between L1 and L2.

    1. A Perpendicular
    2. B The same line (coincident)
    3. C Parallel, and not the same line
    4. D Neither parallel nor perpendicular
  2. 21 mark

    A circle has a diameter with endpoints A(-1, 2) and B(7, 8). Find the equation of the circle, in the form (x - a)^2 + (y - b)^2 = r^2.

    1. A (x - 3)^2 + (y - 5)^2 = 25
    2. B (x - 6)^2 + (y - 10)^2 = 25
    3. C (x - 3)^2 + (y - 5)^2 = 100
    4. D (x - 3)^2 + (y - 5)^2 = 10
  3. 31 mark

    A circle has centre O and radius 13 cm. A chord AB lies at a perpendicular distance of 5 cm from O. Find the length of AB.

    1. A 12 cm
    2. B 18 cm
    3. C 8 cm
    4. D 24 cm
  4. 41 mark

    PT is a tangent to a circle at the point T, and TQ is a chord of the circle. The angle between the tangent PT and the chord TQ is 50 degrees. R is a point on the major arc of the circle. Find the size of angle TRQ, the angle in the alternate segment.

    1. A 100 degrees
    2. B 50 degrees
    3. C 25 degrees
    4. D 130 degrees
  5. 51 mark

    In triangle ABC, AB = 7 cm, AC = 8 cm and angle BAC = 60 degrees. Find the length BC, giving your answer in exact (surd) form where necessary.

    1. A sqrt(57) cm
    2. B sqrt(113) cm
    3. C 13 cm
    4. D sqrt(85) cm
  6. 61 mark

    In triangle ABC, angle A = 45 degrees, angle B = 45 degrees and angle C = 90 degrees. Side AB (opposite angle C, the hypotenuse) has length 10 cm. Use the sine rule to find the length of side BC (opposite angle A), in exact (surd) form.

    1. A 10sqrt(2) cm
    2. B 5 cm
    3. C 5sqrt(2) cm
    4. D 5sqrt(3) cm
  7. 71 mark

    A sector of a circle has radius 8 cm and its arc subtends an angle of pi/3 radians at the centre. Find the exact area of the corresponding segment (the region enclosed between the chord and the arc).

    1. A (32pi/3 + 16sqrt(3)) cm^2
    2. B 16sqrt(3) cm^2
    3. C 32pi/3 cm^2
    4. D (32pi/3 - 16sqrt(3)) cm^2
  8. 81 mark

    Find the exact value of tan(60 degrees) - tan(30 degrees).

    1. A 0
    2. B 2sqrt(3)/3
    3. C 4sqrt(3)/3
    4. D -2sqrt(3)/3
  9. 91 mark

    The graph of y = sin(x) has a maximum at x = 90 degrees, and is symmetric about this maximum, so sin(x) = sin(180 - x) for all x. Given that sin(70 degrees) = sin(x) for some x between 0 and 180 degrees other than 70, find x.

    1. A 20 degrees
    2. B 160 degrees
    3. C 110 degrees
    4. D 250 degrees
  10. 101 mark

    Given that cos(theta) = -2/3 and theta is obtuse (90 < theta < 180 degrees), find the exact value of sin(theta).

    1. A sqrt(5)/3
    2. B -sqrt(5)/3
    3. C sqrt(5)/9
    4. D sqrt(15)/3
  11. 111 mark

    Solve tan(x) = -1 for -180 <= x <= 180 degrees, giving all solutions.

    1. A x = -60 degrees or x = 120 degrees
    2. B x = -45 degrees only
    3. C x = 45 degrees or x = -135 degrees
    4. D x = -45 degrees or x = 135 degrees
  12. 121 mark

    The lines kx - 4y + 1 = 0 and 3x + 6y - 2 = 0 are perpendicular. Find the value of k.

    1. A k = -2
    2. B k = 8
    3. C k = 2
    4. D k = -8
  13. 131 mark

    A circle has centre (1, -3) and passes through the point (4, 1). Find the equation of the circle, in the form (x - a)^2 + (y - b)^2 = r^2.

    1. A (x + 1)^2 + (y - 3)^2 = 25
    2. B (x - 1)^2 + (y + 3)^2 = 17
    3. C (x - 1)^2 + (y + 3)^2 = 25
    4. D (x - 1)^2 + (y + 3)^2 = 7
  14. 141 mark

    ABCD is a cyclic quadrilateral, with vertices in order round the circle. Angle BAD = 85 degrees. Find angle BCD.

    1. A 85 degrees
    2. B 42.5 degrees
    3. C 275 degrees
    4. D 95 degrees
  15. 151 mark

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

    1. A x = 45, 135, 225 or 315 degrees
    2. B x = 45 or 135 degrees only
    3. C x = 30, 150, 210 or 330 degrees
    4. D x = 45 or 225 degrees only

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: C

    1. Find the gradient of L1 using the two given points: m = (10 - 4)/(3 - 0) = 6/3 = 2.
    2. Rearrange L2 into gradient form: 4x - 2y + 7 = 0 gives 2y = 4x + 7, so y = 2x + 3.5, and L2 has gradient 2.
    3. Both lines have gradient 2, so they are parallel. L1 through (0, 4) with gradient 2 is y = 2x + 4, which has a different y-intercept from L2's y = 2x + 3.5, so they are not the same line.
    4. So L1 and L2 are parallel, and not the same line, option C.
    • Why not A: Correctly finds L1's gradient as 2, but then applies the negative-reciprocal rule meant for perpendicular lines, rather than checking whether the two gradients are simply equal.
    • Why not B: Correctly finds that both lines have gradient 2, but stops there and assumes equal gradients means identical lines, without checking that they have different y-intercepts (4 and 3.5).
    • Why not D: Makes an arithmetic slip finding L1's gradient, computing (10 - 0)/(4 - 3) = 10 instead of (10 - 4)/(3 - 0) = 2, and so does not notice that the two gradients actually match.

  2. Question 2Answer: A

    1. The centre of the circle is the midpoint of the diameter: midpoint = ((-1 + 7)/2, (2 + 8)/2) = (3, 5).
    2. The radius is half the length of the diameter. The diameter's length is sqrt((7 - (-1))^2 + (8 - 2)^2) = sqrt(8^2 + 6^2) = sqrt(64 + 36) = sqrt(100) = 10, so the radius is 5.
    3. So r^2 = 25, and the equation is (x - 3)^2 + (y - 5)^2 = 25, option A.
    • Why not B: Finds the radius correctly, but finds the centre by adding the endpoints' coordinates, (-1 + 7, 2 + 8) = (6, 10), instead of averaging them for the midpoint.
    • Why not C: Finds the correct centre (3, 5) by averaging the endpoints, but uses the full length of the diameter, 10, as the radius, forgetting to halve it, so states r^2 = 100 instead of 25.
    • Why not D: Finds the correct centre and correctly identifies the diameter as 10, but writes 10 itself as r^2 instead of squaring the radius (half of 10, which is 5, giving r^2 = 25).

  3. Question 3Answer: D

    1. The perpendicular from the centre of a circle to a chord bisects the chord, so it splits AB into two equal halves, each forming a right-angled triangle with the radius as the hypotenuse.
    2. By Pythagoras' theorem, half of AB = sqrt(radius^2 - distance^2) = sqrt(13^2 - 5^2) = sqrt(169 - 25) = sqrt(144) = 12.
    3. Since this is only half the chord, the full length AB = 2 x 12 = 24 cm, option D.
    • Why not A: Correctly uses Pythagoras' theorem on the right-angled triangle formed by the radius, the perpendicular distance and half the chord, finding half of AB as sqrt(13^2 - 5^2) = sqrt(144) = 12, but forgets that the perpendicular from the centre bisects the chord, so this is only half of AB, not the whole length.
    • Why not B: Adds the radius and the perpendicular distance directly, 13 + 5 = 18, instead of applying Pythagoras' theorem to the right-angled triangle they form with half the chord.
    • Why not C: Subtracts the perpendicular distance from the radius directly, 13 - 5 = 8, instead of applying Pythagoras' theorem.

  4. Question 4Answer: B

    1. The alternate segment theorem states that the angle between a tangent and a chord, at the point of contact, equals the angle in the alternate segment (the segment on the other side of the chord from where the tangent-chord angle is measured).
    2. Here the tangent-chord angle PTQ = 50 degrees, and R lies in the alternate segment (the major arc, on the far side of chord TQ from the tangent-chord angle).
    3. So angle TRQ = 50 degrees, option B, directly by the alternate segment theorem.
    • Why not A: Confuses the alternate segment theorem with the rule that the angle at the centre is twice the angle at the circumference, and doubles the given angle instead of using it directly.
    • Why not C: Confuses the alternate segment theorem with the same centre-circumference rule the other way round, halving the given angle instead of using it directly.
    • Why not D: Confuses the tangent-chord angle with an angle in a cyclic quadrilateral, and subtracts it from 180 degrees instead of applying the alternate segment theorem directly.

  5. Question 5Answer: A

    1. The cosine rule states BC^2 = AB^2 + AC^2 - 2 x AB x AC x cos(BAC).
    2. Substitute the values: BC^2 = 7^2 + 8^2 - 2(7)(8)cos(60) = 49 + 64 - 112 x (1/2) = 113 - 56 = 57.
    3. So BC = sqrt(57) cm, option A.
    • Why not B: Forgets the -2 x AB x AC x cos(BAC) term in the cosine rule entirely, effectively applying Pythagoras' theorem as though angle BAC were 90 degrees, giving BC^2 = 7^2 + 8^2 = 113.
    • Why not C: Uses a + sign instead of a - sign in the cosine rule, computing BC^2 = 7^2 + 8^2 + 2(7)(8)cos(60) = 49 + 64 + 56 = 169, giving BC = 13.
    • Why not D: Forgets to double the product of the two sides in the cosine rule, using BC^2 = 7^2 + 8^2 - (7)(8)cos(60) = 113 - 28 = 85 instead of subtracting 2(7)(8)cos(60) = 56.

  6. Question 6Answer: C

    1. The sine rule states BC/sin(A) = AB/sin(C).
    2. Substitute the known values: BC/sin(45) = 10/sin(90).
    3. sin(90) = 1 and sin(45) = sqrt(2)/2, so BC = 10 x (sqrt(2)/2)/1 = 5sqrt(2) cm, option C.
    • Why not A: Uses sin(45) = sqrt(2) in the sine rule calculation instead of the correct value sqrt(2)/2, giving BC = 10 x sqrt(2)/sin(90) = 10sqrt(2) instead of 10 x (sqrt(2)/2).
    • Why not B: Confuses sin(45) with sin(30) = 1/2, giving BC = 10 x (1/2)/sin(90) = 5.
    • Why not D: Confuses sin(45) with sin(60) = sqrt(3)/2, giving BC = 10 x (sqrt(3)/2)/sin(90) = 5sqrt(3).

  7. Question 7Answer: D

    1. The area of the sector is (1/2) r^2 theta = (1/2)(8^2)(pi/3) = (1/2)(64)(pi/3) = 32pi/3 cm^2.
    2. The area of the triangle formed by the two radii and the chord is (1/2) r^2 sin(theta) = (1/2)(64) sin(pi/3) = 32 x (sqrt(3)/2) = 16sqrt(3) cm^2.
    3. The segment is the sector with the triangle removed: area = 32pi/3 - 16sqrt(3) cm^2, option D.
    • Why not A: Adds the triangle's area to the sector's area instead of subtracting it, rather than removing the triangle to leave just the segment.
    • Why not B: Computes only the area of the triangle formed by the two radii and the chord, (1/2) r^2 sin(theta) = (1/2)(64)sin(pi/3) = 16sqrt(3), forgetting to include the sector at all.
    • Why not C: Computes only the area of the sector, (1/2) r^2 theta = (1/2)(64)(pi/3) = 32pi/3, forgetting to subtract the triangle to leave just the segment.

  8. Question 8Answer: B

    1. Use the exact values tan(60) = sqrt(3) and tan(30) = 1/sqrt(3) = sqrt(3)/3.
    2. Subtract: sqrt(3) - sqrt(3)/3 = (3sqrt(3) - sqrt(3))/3 = 2sqrt(3)/3, option B.
    • Why not A: Assumes tan(60) and tan(30) are equal, perhaps because 60 and 30 are complementary angles, so treats the subtraction as giving 0.
    • Why not C: Adds the two exact values instead of subtracting them: sqrt(3) + sqrt(3)/3 = 4sqrt(3)/3.
    • Why not D: Swaps the two exact values, using tan(30) = sqrt(3) and tan(60) = sqrt(3)/3 the wrong way round, giving sqrt(3)/3 - sqrt(3) = -2sqrt(3)/3.

  9. Question 9Answer: C

    1. The graph of y = sin(x) is symmetric about its maximum at x = 90 degrees, which means sin(x) = sin(180 - x) for all x.
    2. So the other solution to sin(x) = sin(70 degrees) in the range 0 to 180 degrees is x = 180 - 70 = 110 degrees, option C.
    • Why not A: Reflects the angle about x = 90 by subtracting it from 90 rather than from 180, computing 90 - 70 = 20 instead of 180 - 70.
    • Why not B: Adds 90 degrees to the original angle instead of reflecting it about 180 degrees, computing 70 + 90 = 160.
    • Why not D: Adds 180 degrees to the original angle instead of subtracting it from 180 degrees, computing 70 + 180 = 250, which also falls outside the stated range of 0 to 180 degrees.

  10. Question 10Answer: A

    1. Use sin^2(theta) + cos^2(theta) = 1: sin^2(theta) = 1 - cos^2(theta) = 1 - (-2/3)^2 = 1 - 4/9 = 5/9.
    2. Since theta is obtuse (between 90 and 180 degrees), sin(theta) is positive.
    3. So sin(theta) = sqrt(5/9) = sqrt(5)/3, option A.
    • Why not B: Correctly finds sin^2(theta) = 5/9, but wrongly takes the negative square root, reasoning that since cos(theta) is negative sin(theta) must be too, rather than recalling that sine is positive throughout the second quadrant (90 to 180 degrees).
    • Why not C: Correctly finds sin^2(theta) = 1 - (2/3)^2 = 5/9, but when taking the square root only applies it to the numerator, giving sqrt(5)/9 instead of sqrt(5)/3.
    • Why not D: Uses the identity as sin^2(theta) = 1 - cos(theta) instead of 1 - cos^2(theta), forgetting to square cos(theta) first, giving sin^2(theta) = 1 - (-2/3) = 5/3 and so sin(theta) = sqrt(5/3) = sqrt(15)/3.

  11. Question 11Answer: D

    1. Since tan(45) = 1, the reference angle is 45 degrees.
    2. tan(x) is negative in the second and fourth quadrants.
    3. In the range -180 to 180 degrees, the fourth-quadrant solution is x = -45 degrees, and the second-quadrant solution is x = 180 - 45 = 135 degrees.
    4. So the solutions are x = -45 degrees or x = 135 degrees, option D.
    • Why not A: Uses a reference angle of 60 degrees, confusing tan(60) = sqrt(3) with the required tan(45) = 1, instead of using the correct reference angle of 45 degrees.
    • Why not B: Finds one valid solution but forgets that tan has period 180 degrees, so misses the second solution that also lies within the given range.
    • Why not C: Forgets that tan(x) = -1 is negative, and instead places the solutions in the first and third quadrants, where tangent is positive, rather than the second and fourth quadrants, where tangent is negative.

  12. Question 12Answer: B

    1. Rearrange 3x + 6y - 2 = 0 into gradient form: 6y = -3x + 2, so y = -(1/2)x + 1/3, giving gradient -1/2.
    2. Rearrange kx - 4y + 1 = 0 into gradient form: -4y = -kx - 1, so y = (k/4)x + 1/4, giving gradient k/4.
    3. For perpendicular lines, the product of the gradients equals -1: (k/4)(-1/2) = -1, so -k/8 = -1, giving k = 8, option B.
    • Why not A: Sets the two gradients equal to each other, the condition for parallel lines, instead of setting their product equal to -1, the condition for perpendicular lines.
    • Why not C: Reads the gradient of kx - 4y + 1 = 0 directly as k, the coefficient of x, instead of rearranging to y = (k/4)x + 1/4 and using k/4 as the gradient.
    • Why not D: Sets the product of the two gradients equal to +1 instead of -1, missing the negative sign that the perpendicular condition requires.

  13. Question 13Answer: C

    1. The radius is the distance from the centre (1, -3) to the point (4, 1) on the circle: r^2 = (4 - 1)^2 + (1 - (-3))^2 = 3^2 + 4^2 = 9 + 16 = 25.
    2. So the equation is (x - 1)^2 + (y + 3)^2 = 25, option C.
    • Why not A: Flips the signs of the centre's coordinates when writing the equation, using (x + 1)^2 + (y - 3)^2 instead of (x - 1)^2 + (y + 3)^2 for a centre of (1, -3).
    • Why not B: Treats the given point's coordinates as distances from the origin rather than from the centre, computing r^2 = 4^2 + 1^2 = 17 instead of subtracting the centre's coordinates first.
    • Why not D: Adds the coordinate differences directly instead of squaring and using Pythagoras, computing r^2 = (4 - 1) + (1 - (-3)) = 3 + 4 = 7 instead of 3^2 + 4^2.

  14. Question 14Answer: D

    1. Opposite angles of a cyclic quadrilateral sum to 180 degrees.
    2. Angle BAD and angle BCD are opposite angles in ABCD, so angle BCD = 180 - 85 = 95 degrees, option D.
    • Why not A: Assumes opposite angles of a cyclic quadrilateral are equal, rather than recalling that they are supplementary (sum to 180 degrees).
    • Why not B: Confuses the cyclic quadrilateral rule with the centre-circumference angle theorem, and halves the given angle instead of subtracting it from 180 degrees.
    • Why not C: Subtracts the given angle from a full turn of 360 degrees instead of from the straight angle of 180 degrees that the cyclic quadrilateral rule actually uses.

  15. Question 15Answer: A

    1. Divide by 2: sin^2(x) = 1/2, so sin(x) = sqrt(2)/2 or sin(x) = -sqrt(2)/2.
    2. Since sin(45) = sqrt(2)/2, the reference angle is 45 degrees.
    3. For sin(x) = sqrt(2)/2 (positive, so first and second quadrants): x = 45 degrees or x = 180 - 45 = 135 degrees.
    4. For sin(x) = -sqrt(2)/2 (negative, so third and fourth quadrants): x = 180 + 45 = 225 degrees or x = 360 - 45 = 315 degrees.
    5. So the solutions are x = 45, 135, 225 or 315 degrees, option A.
    • Why not B: Correctly finds sin^2(x) = 1/2, but only takes the positive square root sin(x) = sqrt(2)/2, forgetting that sin(x) = -sqrt(2)/2 is also a valid solution, so misses the two solutions where sine is negative.
    • Why not C: Forgets to take the square root when solving sin^2(x) = 1/2, treating the equation as though it read sin(x) = 1/2 directly, and so uses the wrong reference angle of 30 degrees instead of 45 degrees.
    • Why not D: Finds one solution for each sign of sin(x) but forgets that each sign gives two solutions in the range 0 to 360 degrees, missing the supplementary-angle partner of each one (135 and 315 degrees).

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