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

Test standard. 15 questions, 15 marks, about 22 minutes.

ESAT Mathematics 2: Coordinate geometry and trigonometry, set 1

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

    Find the equation of the straight line passing through the points (2, 3) and (5, 12).

    1. A y = 3x - 3
    2. B y = (1/3)x + 7/3
    3. C y = 3x + 1
    4. D y = -3x + 9
  2. 21 mark

    Line L1 has equation 3x - 4y + 8 = 0. Line L2 is perpendicular to L1 and passes through the point (4, -1). Find the equation of L2 in the form ax + by + c = 0.

    1. A 3x - 4y - 16 = 0
    2. B 4x - 3y - 19 = 0
    3. C 4x + 3y - 13 = 0
    4. D 4x + 3y + 13 = 0
  3. 31 mark

    A circle has equation (x - 3)^2 + (y + 2)^2 = 25. State the centre and radius of the circle.

    1. A Centre (-3, 2), radius 5
    2. B Centre (3, -2), radius 5
    3. C Centre (3, -2), radius 25
    4. D Centre (-3, 2), radius 25
  4. 41 mark

    A circle has equation x^2 + y^2 - 6x + 4y - 3 = 0. Find the radius of the circle.

    1. A 16
    2. B sqrt(3)
    3. C sqrt(10)
    4. D 4
  5. 51 mark

    A circle has equation x^2 + y^2 = 25. Find the equation of the tangent to the circle at the point (3, 4), in the form ax + by = c.

    1. A 4x - 3y = 0
    2. B 3x + 4y = 25
    3. C 3x - 4y = -7
    4. D 3x + 4y = 24
  6. 61 mark

    Points A, B and C lie on a circle with centre O. The angle AOB at the centre is 140 degrees, and C lies on the major arc AB. Find the size of angle ACB.

    1. A 70 degrees
    2. B 140 degrees
    3. C 40 degrees
    4. D 110 degrees
  7. 71 mark

    In triangle ABC, angle A = 30 degrees, side c = 10 cm (the side adjacent to A, i.e. AB) and side a = 6 cm (the side opposite A, i.e. BC). How many distinct triangles ABC satisfy these measurements?

    1. A 1 triangle
    2. B 0 triangles
    3. C Infinitely many triangles
    4. D 2 triangles
  8. 81 mark

    Triangle PQR has PQ = 6 cm, PR = 5 cm and angle QPR = 60 degrees. Find the area of triangle PQR, giving your answer in exact (surd) form.

    1. A 15sqrt(3) cm^2
    2. B (11sqrt(3))/4 cm^2
    3. C (15sqrt(3))/2 cm^2
    4. D 15/2 cm^2
  9. 91 mark

    Given that sin(theta) = 3/5 and theta is acute (0 < theta < 90 degrees), find the exact value of tan(theta).

    1. A 3/4
    2. B 4/3
    3. C -3/4
    4. D 15/16
  10. 101 mark

    Solve cos(x) = -1/2 for 0 <= x <= 360 degrees, giving all solutions.

    1. A x = 120 degrees only
    2. B x = 150 degrees or x = 210 degrees
    3. C x = 240 degrees or x = 300 degrees
    4. D x = 120 degrees or x = 240 degrees
  11. 111 mark

    A sector of a circle has radius 6 cm, and its arc subtends an angle of (2pi)/3 radians at the centre. Find the length of the arc.

    1. A 12pi cm
    2. B 4pi cm
    3. C (2pi)/3 cm
    4. D 8pi cm
  12. 121 mark

    Find the exact value of 2 sin(60 degrees) cos(30 degrees).

    1. A sqrt(3)/2
    2. B 3/4
    3. C 3/2
    4. D 1
  13. 131 mark

    The graph of y = cos(x) has a line of symmetry at x = 180 degrees. Using the periodicity and symmetry of the cosine graph, find the value of x in the range 360 < x < 720 degrees at which the graph has a line of symmetry.

    1. A x = 360 degrees
    2. B x = 450 degrees
    3. C x = 720 degrees
    4. D x = 540 degrees
  14. 141 mark

    Simplify fully: sin(theta) / tan(theta).

    1. A cos(theta)
    2. B sin^2(theta) / cos(theta)
    3. C sin(theta) cos(theta)
    4. D 1 / cos(theta)
  15. 151 mark

    Solve sin(x) = sqrt(3)/2 for 0 <= x <= 360 degrees, giving all solutions.

    1. A x = 60 degrees only
    2. B x = 60 degrees or x = 120 degrees
    3. C x = 240 degrees or x = 300 degrees
    4. D x = 30 degrees or x = 150 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: A

    1. Find the gradient between the two given points: m = (12 - 3)/(5 - 2) = 9/3 = 3.
    2. Use the point-gradient form with (2, 3): y - 3 = 3(x - 2).
    3. Expand and simplify: y - 3 = 3x - 6, so y = 3x - 3.
    4. Check with the other point (5, 12): 3(5) - 3 = 12, which matches, so the equation is y = 3x - 3, option A.
    • Why not B: Finds the gradient using run over rise instead of rise over run, computing (5 - 2)/(12 - 3) = 3/9 = 1/3 instead of (12 - 3)/(5 - 2) = 3.
    • Why not C: Uses the correct gradient of 3 but expands the point-gradient form incorrectly, treating y - 3 = 3(x - 2) as y - 3 = 3x - 2 (forgetting to multiply the 2 by 3 as well), giving y = 3x + 1.
    • Why not D: Computes the gradient with the y-coordinates taken in the wrong order relative to the x-coordinates, using (3 - 12)/(5 - 2) = -3 instead of (12 - 3)/(5 - 2) = 3.

  2. Question 2Answer: C

    1. Rearrange L1 into gradient form: 3x - 4y + 8 = 0 gives 4y = 3x + 8, so y = (3/4)x + 2, and L1 has gradient 3/4.
    2. A line perpendicular to L1 has gradient equal to the negative reciprocal of 3/4, which is -4/3.
    3. Use the point (4, -1) with this gradient: y - (-1) = -(4/3)(x - 4), so y + 1 = -(4/3)x + 16/3.
    4. Multiply through by 3 and rearrange: 3y + 3 = -4x + 16, so 4x + 3y - 13 = 0, which is option C.
    • Why not A: Rearranges L1 to find its gradient (3/4) correctly, but then uses this same gradient for L2 instead of its negative reciprocal, effectively treating 'perpendicular' as 'parallel'.
    • Why not B: Correctly finds L1's gradient (3/4) and takes its reciprocal (4/3) to build a perpendicular-looking gradient, but forgets to also change its sign, using 4/3 instead of -4/3.
    • Why not D: Uses the correct perpendicular gradient of -4/3, but substitutes the given point with its coordinates' signs swapped, using (-4, 1) instead of (4, -1).

  3. Question 3Answer: B

    1. The equation of a circle with centre (a, b) and radius r is (x - a)^2 + (y - b)^2 = r^2.
    2. Comparing (x - 3)^2 + (y + 2)^2 = 25 to this form: x - a = x - 3 gives a = 3, and y - b = y + 2 = y - (-2) gives b = -2, so the centre is (3, -2).
    3. The right-hand side is r^2 = 25, so r = sqrt(25) = 5.
    4. So the circle has centre (3, -2) and radius 5, option B.
    • Why not A: Reads the centre coordinates straight off the numbers inside the brackets without changing their signs, giving (-3, 2) instead of (3, -2); since the equation is (x - a)^2 + (y - b)^2 = r^2, the centre is actually (3, -2).
    • Why not C: Correctly reads the centre as (3, -2) but takes the number on the right-hand side, 25, as the radius directly, forgetting that this is r^2, so the radius is sqrt(25) = 5, not 25.
    • Why not D: Combines both errors: reads the centre with the signs unchanged as (-3, 2), and also takes 25 itself as the radius instead of taking its square root.

  4. Question 4Answer: D

    1. Compare x^2 + y^2 - 6x + 4y - 3 = 0 with the general form x^2 + y^2 + cx + dy + e = 0: here c = -6, d = 4, e = -3.
    2. Complete the square: x^2 - 6x = (x - 3)^2 - 9, and y^2 + 4y = (y + 2)^2 - 4.
    3. Substitute back: (x - 3)^2 - 9 + (y + 2)^2 - 4 - 3 = 0, so (x - 3)^2 + (y + 2)^2 = 16.
    4. This matches (x - a)^2 + (y - b)^2 = r^2 with r^2 = 16, so the radius r = sqrt(16) = 4, option D.
    • Why not A: Correctly completes the square to find radius^2 = 16, but then states this squared value itself as the radius, forgetting to take the square root.
    • Why not B: Uses radius^2 = -(constant term) only, giving radius^2 = -(-3) = 3, and omits the (c/2)^2 + (d/2)^2 contribution that comes from the x and y coefficients.
    • Why not C: Uses the correct terms from the x and y coefficients, (c/2)^2 + (d/2)^2 = 9 + 4 = 13, but adds the constant term instead of subtracting it, giving radius^2 = 13 + (-3) = 10 instead of 13 - (-3) = 16.

  5. Question 5Answer: B

    1. The point (3, 4) lies on the circle since 3^2 + 4^2 = 9 + 16 = 25.
    2. The radius from the centre (0, 0) to (3, 4) has gradient 4/3.
    3. The tangent at (3, 4) is perpendicular to this radius, so its gradient is the negative reciprocal: -3/4.
    4. Using the point (3, 4): y - 4 = -(3/4)(x - 3), so y = -(3/4)x + 9/4 + 4 = -(3/4)x + 25/4.
    5. Multiply through by 4 and rearrange: 4y = -3x + 25, so 3x + 4y = 25, option B.
    • Why not A: Uses the gradient of the radius to (3, 4), which is 4/3, as the gradient of the tangent itself, forgetting that a tangent is perpendicular to the radius at the point of contact rather than parallel to it.
    • Why not C: Correctly takes the reciprocal of the radius's gradient (4/3), getting 3/4, but forgets to also change its sign, using 3/4 instead of the correct perpendicular gradient of -3/4.
    • Why not D: Uses the correct perpendicular gradient of -3/4, but substitutes the given point with its coordinates swapped, using (4, 3) instead of (3, 4).

  6. Question 6Answer: A

    1. The angle at the centre of a circle is twice the angle at the circumference, when both angles are subtended by the same arc AB and the circumference point lies on the other (major) arc.
    2. Here C lies on the major arc AB, so angle ACB is subtended by the same arc AB as the given centre angle AOB = 140 degrees.
    3. Applying the theorem: angle ACB = (1/2) x angle AOB = (1/2) x 140 = 70 degrees, option A.
    • Why not B: Forgets that the angle at the centre is twice the angle at the circumference on the same arc, and simply copies the given centre angle as the circumference angle.
    • Why not C: Wrongly applies the cyclic quadrilateral rule (opposite angles sum to 180 degrees) to the centre and circumference angles, computing 180 - 140 = 40 instead of using the correct centre-circumference relationship.
    • Why not D: Uses the reflex angle at O, 360 - 140 = 220, instead of the given (non-reflex) angle AOB = 140, and then halves this wrong value, confusing which arc corresponds to C's position on the major arc.

  7. Question 7Answer: D

    1. This is an angle-side-side (SSA) situation: angle A and side a are given, along with the adjacent side c, which is exactly the setup where the ambiguous case can arise.
    2. Find the perpendicular height from B to the line AC: h = c sin A = 10 x sin 30 = 10 x (1/2) = 5.
    3. Compare h, a and c: h = 5, a = 6 and c = 10, so h < a < c.
    4. When h < a < c, the side of length a can swing to meet the base line at two different points, giving two distinct triangles, so the answer is 2, option D.
    • Why not A: Treats this angle-side-side information the same way as angle-side-angle or side-angle-side data, assuming it fixes a unique triangle, and misses that two different configurations can satisfy the same angle and side lengths.
    • Why not B: Compares the given side a directly to the full adjacent side c rather than to the perpendicular height h = c sin A, and wrongly concludes that side a is too short to form a triangle.
    • Why not C: Misunderstands the ambiguous case as meaning the triangle is left completely undetermined, rather than as meaning exactly two distinct valid triangles both satisfy the given data.

  8. Question 8Answer: C

    1. The area of a triangle with two sides a, b and the included angle C is (1/2) ab sin C.
    2. Here a = PQ = 6, b = PR = 5 and the included angle C = 60 degrees, so area = (1/2)(6)(5) sin(60).
    3. sin(60) = sqrt(3)/2, so area = (1/2)(30)(sqrt(3)/2) = 15sqrt(3)/2, option C.
    • Why not A: Uses the correct product PQ x PR x sin(60) = 30 x (sqrt(3)/2) = 15sqrt(3), but forgets to include the leading factor of 1/2 in the area formula (1/2)ab sin C.
    • Why not B: Adds the two given sides instead of multiplying them, computing (1/2)(PQ + PR) sin(60) = (1/2)(11)(sqrt(3)/2) instead of (1/2)(PQ)(PR) sin(60).
    • Why not D: Confuses sin(60) with its complementary value cos(60) = 1/2 (equivalently, with sin(30) = 1/2), using 1/2 in place of sqrt(3)/2 in the area formula.

  9. Question 9Answer: A

    1. Use sin^2(theta) + cos^2(theta) = 1 to find cos(theta): cos^2(theta) = 1 - (3/5)^2 = 1 - 9/25 = 16/25.
    2. Since theta is acute, cos(theta) is positive, so cos(theta) = sqrt(16/25) = 4/5.
    3. Use tan(theta) = sin(theta)/cos(theta) = (3/5)/(4/5) = 3/4, option A.
    • Why not B: Finds cos(theta) correctly as 4/5, but then computes tan(theta) as cos(theta)/sin(theta) instead of sin(theta)/cos(theta).
    • Why not C: Takes the negative square root when finding cos(theta) from cos^2(theta) = 1 - (3/5)^2 = 16/25, giving cos(theta) = -4/5, even though theta is stated to be acute and cosine must be positive in that range.
    • Why not D: Finds cos^2(theta) = 1 - (3/5)^2 = 16/25 correctly, but then uses this squared value directly as cos(theta) itself, forgetting to take the square root before dividing.

  10. Question 10Answer: D

    1. The reference angle is found from cos(reference) = 1/2, and since cos(60) = 1/2, the reference angle is 60 degrees.
    2. Cosine is negative in the second and third quadrants (between 90 and 270 degrees).
    3. In the second quadrant: x = 180 - 60 = 120 degrees. In the third quadrant: x = 180 + 60 = 240 degrees.
    4. So the solutions in the range 0 to 360 degrees are x = 120 degrees or x = 240 degrees, option D.
    • Why not A: Finds one valid solution but forgets that cosine is negative in two quadrants within the given range, missing the second solution.
    • Why not B: Confuses the reference angle: recalls cos(30) = 1/2 instead of cos(60) = 1/2, and uses a reference angle of 30 degrees instead of 60 degrees, giving 180 - 30 = 150 and 180 + 30 = 210.
    • Why not C: Uses the quadrants where sine (rather than cosine) is negative, the third and fourth quadrants, instead of the quadrants where cosine is negative, the second and third.

  11. Question 11Answer: B

    1. The formula for arc length, when the angle theta is in radians, is arc length = r theta.
    2. Here r = 6 and theta = (2pi)/3, so arc length = 6 x (2pi/3).
    3. 6 x (2pi/3) = (12pi)/3 = 4pi, so the arc length is 4pi cm, option B.
    • Why not A: Uses the formula for the area of a sector, (1/2) r^2 theta, instead of the formula for arc length, r theta, computing (1/2)(36)(2pi/3) = 12pi.
    • Why not C: Uses arc length = theta directly, forgetting to multiply by the radius, giving just (2pi)/3.
    • Why not D: Uses the diameter (12 cm) in place of the radius in the arc length formula, computing 12 x (2pi/3) = 8pi instead of 6 x (2pi/3).

  12. Question 12Answer: C

    1. Use the exact values sin(60) = sqrt(3)/2 and cos(30) = sqrt(3)/2.
    2. Substitute: 2 sin(60) cos(30) = 2 x (sqrt(3)/2) x (sqrt(3)/2).
    3. Multiply the surds first: (sqrt(3)/2) x (sqrt(3)/2) = 3/4, so the expression becomes 2 x (3/4) = 3/2, option C.
    • Why not A: Confuses sin(60) = sqrt(3)/2 with sin(30) = 1/2, using 2 x (1/2) x cos(30) = 2 x (1/2) x (sqrt(3)/2) = sqrt(3)/2.
    • Why not B: Correctly finds sin(60) cos(30) = (sqrt(3)/2)(sqrt(3)/2) = 3/4, but forgets to multiply by the leading factor of 2 in the expression.
    • Why not D: Confuses cos(30) = sqrt(3)/2 with tan(30) = 1/sqrt(3), computing 2 sin(60) tan(30) = 2 x (sqrt(3)/2) x (1/sqrt(3)) = 1.

  13. Question 13Answer: D

    1. The cosine graph has lines of symmetry at each of its turning points (maxima and minima), which occur every 180 degrees: at x = 0, 180, 360, 540, 720, and so on.
    2. The range required is 360 < x < 720, which is an open interval excluding both endpoints.
    3. Of the symmetry lines 360, 540 and 720, only x = 540 lies strictly inside this open range, option D.
    • Why not A: Correctly identifies that x = 360 is a line of symmetry of the cosine graph, but forgets that the given range strictly excludes 360 itself.
    • Why not B: Confuses the graph's lines of symmetry, which occur at its turning points (its maxima and minima), with its x-intercepts, which occur 90 degrees away from them.
    • Why not C: Correctly identifies that x = 720 is a line of symmetry of the cosine graph, but forgets that the given range strictly excludes 720 itself.

  14. Question 14Answer: A

    1. Rewrite tan(theta) using the identity tan(theta) = sin(theta)/cos(theta).
    2. So sin(theta) / tan(theta) = sin(theta) / [sin(theta)/cos(theta)].
    3. Dividing by a fraction means multiplying by its reciprocal: sin(theta) x [cos(theta)/sin(theta)].
    4. The sin(theta) terms cancel, leaving cos(theta), option A.
    • Why not B: Multiplies sin(theta) by tan(theta) instead of dividing by it, computing sin(theta) x [sin(theta)/cos(theta)] = sin^2(theta)/cos(theta).
    • Why not C: Misremembers tan(theta) as 1/cos(theta) (confusing it with sec(theta)) rather than sin(theta)/cos(theta), giving sin(theta) / [1/cos(theta)] = sin(theta) cos(theta).
    • Why not D: Divides by the fraction sin(theta)/cos(theta) by cancelling the sin(theta) terms directly rather than inverting and multiplying, incorrectly leaving 1/cos(theta).

  15. Question 15Answer: B

    1. Since sin(60) = sqrt(3)/2, the reference angle is 60 degrees.
    2. Sine is positive in the first and second quadrants (between 0 and 180 degrees).
    3. In the first quadrant: x = 60 degrees. In the second quadrant: x = 180 - 60 = 120 degrees.
    4. So the solutions in the range 0 to 360 degrees are x = 60 degrees or x = 120 degrees, option B.
    • Why not A: Finds the principal solution correctly but forgets that sine is also positive in the second quadrant, missing the second solution.
    • Why not C: Uses the quadrants where sine is negative, the third and fourth quadrants, instead of the quadrants where sine is positive, the first and second.
    • Why not D: Confuses sin(30) = 1/2 with sin(60) = sqrt(3)/2, using a reference angle of 30 degrees instead of 60 degrees.

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