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

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

ESAT Mathematics 2: Coordinate geometry and trigonometry, set 3

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.
  1. 11 mark

    Find the equation of the perpendicular bisector of the line segment joining the points (1, 5) and (7, -3), giving your answer in the form ax + by = c.

    1. A 4x + 3y = 19
    2. B 3x - 4y = 8
    3. C 3x - 4y = -4
    4. D 3x - 4y = 16
  2. 21 mark

    Find the values of k for which the line y = x + k is a tangent to the circle x^2 + y^2 = 18.

    1. A k = 3sqrt(2) or k = -3sqrt(2)
    2. B k = 18sqrt(2) or k = -18sqrt(2)
    3. C k = 6 only
    4. D k = 6 or k = -6
  3. 31 mark

    From an external point P, two tangents PA and PB touch a circle at points A and B, where angle APB = 50 degrees. Point Q lies on the major arc AB. Find the size of angle AQB.

    1. A 65 degrees
    2. B 25 degrees
    3. C 130 degrees
    4. D 115 degrees
  4. 41 mark

    In triangle ABC, angle B = 30 degrees, side b (opposite B) = 6 cm, and side c (opposite C) = 6sqrt(2) cm. This data satisfies the ambiguous (angle-side-side) case for the sine rule. Find the two possible values of angle C.

    1. A C = 45 degrees only
    2. B C = 60 degrees or C = 120 degrees
    3. C C = 45 degrees or C = 135 degrees
    4. D C = 45 degrees or C = 315 degrees
  5. 51 mark

    Triangle ABC has AB = 7 cm, AC = 8 cm and BC = 13 cm. Find the size of angle BAC.

    1. A 60 degrees
    2. B 120 degrees
    3. C 150 degrees
    4. D 180 degrees
  6. 61 mark

    A sector of a circle has radius 6 cm and the angle at the centre is 5pi/6 radians. Find the exact area of the minor segment cut off by the chord.

    1. A 15pi - 9 cm^2
    2. B 15pi cm^2
    3. C 9 cm^2
    4. D 15pi + 9 cm^2
  7. 71 mark

    Lines L1: 2x + y = 8 and L2: x - y = 1 intersect at point P. Line L3 passes through P and is parallel to the line y = 3x - 5. Find the equation of L3.

    1. A y = 3x - 27
    2. B y = -(1/3)x + 3
    3. C y = 3x - 1
    4. D y = 3x - 7
  8. 81 mark

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

    1. A x = 120 degrees or x = 240 degrees
    2. B x = 0 degrees or x = 360 degrees
    3. C x = 0, 120, 240 or 360 degrees
    4. D x = 0, 60, 300 or 360 degrees
  9. 91 mark

    Simplify fully: (1 - sin^2(theta)) / (1 - sin(theta)), given that sin(theta) != 1.

    1. A 1 - sin(theta)
    2. B 1 + sin(theta)
    3. C (1 - sin(theta))^2
    4. D cos(theta)
  10. 101 mark

    The graph of y = tan(x) is sketched for -180 <= x <= 180 degrees. State the number of vertical asymptotes the graph has in this interval, and the values of x at which they occur.

    1. A 4 asymptotes, at x = -180, -90, 90 and 180 degrees
    2. B 1 asymptote, at x = 90 degrees only
    3. C 2 asymptotes, at x = -180 and x = 180 degrees
    4. D 2 asymptotes, at x = -90 and x = 90 degrees
  11. 111 mark

    A chord AB of a circle has length 10 cm. The perpendicular distance from the centre of the circle to the chord is 4 cm. Find the radius of the circle, giving your answer in exact form.

    1. A sqrt(41) cm
    2. B 2sqrt(29) cm
    3. C 3 cm
    4. D 9 cm
  12. 121 mark

    In triangle ABC, angle A = 30 degrees, angle B = 90 degrees, and side a (opposite angle A) = 5 cm. Find the exact area of triangle ABC.

    1. A 25/2 cm^2
    2. B (25sqrt(3))/6 cm^2
    3. C (25sqrt(3))/2 cm^2
    4. D 25sqrt(3) cm^2
  13. 131 mark

    The line y = x - 1 intersects the circle x^2 + y^2 = 25 at two points. Find the exact distance between these two points.

    1. A 7 cm
    2. B 98 cm
    3. C 14 cm
    4. D 7sqrt(2) cm
  14. 141 mark

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

    1. A x = 60 degrees or x = 240 degrees
    2. B x = 60 degrees only
    3. C x = 30 degrees or x = 210 degrees
    4. D x = 120 degrees or x = 300 degrees
  15. 151 mark

    A circle has diameter endpoints A(-4, 1) and B(6, 1). Point C lies on the circle and has y-coordinate 5. Find the possible value(s) of the x-coordinate of C.

    1. A x = 6 or x = -2
    2. B x = 1 only
    3. C x = 4 or x = -2
    4. D x = 4 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: B

    1. The gradient of the segment joining (1, 5) and (7, -3) is (-3 - 5)/(7 - 1) = -8/6 = -4/3.
    2. A perpendicular bisector has gradient equal to the negative reciprocal of this, which is 3/4.
    3. The midpoint of the segment is ((1 + 7)/2, (5 + (-3))/2) = (4, 1), and the bisector passes through it.
    4. Using y - 1 = (3/4)(x - 4): multiplying by 4 gives 4y - 4 = 3x - 12, so 3x - 4y = 8, option B.
    • Why not A: Finds the midpoint (4, 1) correctly, but constructs the line using the original segment's gradient, -4/3, instead of its perpendicular, treating the bisector as parallel to the segment rather than perpendicular to it.
    • Why not C: Uses the correct perpendicular gradient of 3/4, but miscalculates the midpoint's y-coordinate as (5 + 3)/2 = 4 instead of (5 + (-3))/2 = 1, mishandling the negative sign on the second point's y-coordinate.
    • Why not D: Uses the correct perpendicular gradient of 3/4, but forgets to divide by 2 when finding the midpoint, using the coordinate sum (8, 2) directly instead of the average (4, 1).

  2. Question 2Answer: D

    1. Substitute y = x + k into x^2 + y^2 = 18: x^2 + (x + k)^2 = 18, which expands to 2x^2 + 2kx + (k^2 - 18) = 0.
    2. For the line to be a tangent, this quadratic in x must have a repeated root, so its discriminant is zero.
    3. Discriminant: (2k)^2 - 4(2)(k^2 - 18) = 4k^2 - 8k^2 + 144 = -4k^2 + 144. Setting this to zero gives k^2 = 36.
    4. So k = 6 or k = -6, option D. (Check: the distance from the origin to x - y + k = 0 is |k|/sqrt(2); setting this to the radius sqrt(18) = 3sqrt(2) gives |k| = 6, confirming the result.)
    • Why not A: Expands (x + k)^2 as x^2 + k^2, dropping the cross term 2kx, so arrives at 2x^2 + (k^2 - 18) = 0 with no linear term and sets its discriminant to zero, giving k^2 = 18 instead of k^2 = 36.
    • Why not B: Treats the 18 on the right-hand side of the circle's equation as the radius itself rather than as the radius squared, so sets the distance from the origin to the line equal to 18 instead of to sqrt(18), giving k = 18sqrt(2) instead of k = 6.
    • Why not C: Correctly finds that |k| = 6, but gives only the positive value, forgetting that a line can be tangent to the circle on either side, so k = -6 is an equally valid solution.

  3. Question 3Answer: A

    1. OA is perpendicular to the tangent PA, and OB is perpendicular to the tangent PB, since a radius is perpendicular to the tangent at the point of contact.
    2. In quadrilateral OAPB, the interior angles sum to 360 degrees: angle OAP + angle APB + angle PBO + angle BOA = 360, so 90 + 50 + 90 + angle AOB = 360, giving angle AOB = 130 degrees.
    3. Q lies on the major arc AB, on the opposite side of chord AB from the arc that angle AOB directly subtends, so by the angle at the centre theorem, angle AQB is half of this central angle.
    4. angle AQB = 130 / 2 = 65 degrees, option A.
    • Why not B: Assumes the angle AOB at the centre equals the tangent angle APB directly, missing that OAPB is a quadrilateral with right angles at A and B, so angle AOB is actually 180 - 50 = 130 degrees; halving 50 instead gives 25 degrees.
    • Why not C: Correctly finds angle AOB = 180 - 50 = 130 degrees using the quadrilateral OAPB, but then treats this central angle as equal to the angle at the circumference, forgetting to halve it.
    • Why not D: Correctly finds angle AOB = 130 degrees, but halves the reflex angle on Q's own side of the chord (360 - 130 = 230) instead of the non-reflex angle on the opposite side, giving 115 degrees instead of 65.

  4. Question 4Answer: C

    1. By the sine rule, sin(C)/c = sin(B)/b, so sin(C) = c sin(B)/b = 6sqrt(2) x sin(30)/6 = 6sqrt(2) x (1/2)/6 = sqrt(2)/2.
    2. The principal solution is C = 45 degrees, since sin(45) = sqrt(2)/2.
    3. The second possible solution within a triangle is C = 180 - 45 = 135 degrees.
    4. Checking validity: if C = 45, angle A = 180 - 30 - 45 = 105 (valid); if C = 135, angle A = 180 - 30 - 135 = 15 (valid). Both triangles exist, so C = 45 degrees or C = 135 degrees, option C.
    • Why not A: Correctly finds the principal solution C = 45 degrees from sin(C) = sqrt(2)/2, but forgets that the ambiguous case gives a second valid solution, 180 - 45 = 135 degrees.
    • Why not B: Confuses the exact value sin(45 degrees) = sqrt(2)/2 with sin(60 degrees) = sqrt(3)/2, using a reference angle of 60 degrees instead of 45 degrees for both solutions.
    • Why not D: Finds the correct reference angle 45 degrees, but generates the second solution as 360 - 45 instead of 180 - 45, forgetting that the angles of a triangle cannot exceed 180 degrees.

  5. Question 5Answer: B

    1. By the cosine rule, cos(A) = (AB^2 + AC^2 - BC^2)/(2 x AB x AC) = (49 + 64 - 169)/(2 x 7 x 8) = -56/112 = -1/2.
    2. Since cos(A) is negative, angle A is obtuse.
    3. The angle with cos(A) = -1/2 is A = 180 - 60 = 120 degrees, since cos(60) = 1/2, option B.
    • Why not A: Correctly computes cos(A) = -1/2, but ignores the negative sign and finds the acute angle whose cosine has magnitude 1/2, missing that a negative cosine means A is obtuse.
    • Why not C: Correctly recognises that A is obtuse, but recalls the wrong exact value, confusing cos(120) = -1/2 with cos(150) = -sqrt(3)/2.
    • Why not D: Omits the factor of 2 in the denominator of the cosine rule, computing cos(A) = -56/(7 x 8) = -1 instead of -56/(2 x 7 x 8) = -1/2, giving A = 180 degrees.

  6. Question 6Answer: A

    1. The area of the sector is (1/2)r^2(theta) = (1/2)(36)(5pi/6) = 15pi cm^2.
    2. The area of the triangle enclosed by the two radii and the chord is (1/2)r^2 sin(theta) = (1/2)(36) sin(5pi/6). Since sin(5pi/6) = sin(pi - 5pi/6) = sin(pi/6) = 1/2, this is 18 x (1/2) = 9 cm^2.
    3. The segment's area is the sector's area minus the triangle's area: 15pi - 9 cm^2, option A.
    • Why not B: Gives the area of the sector alone, forgetting that the segment is the sector's area with the enclosed triangle's area subtracted.
    • Why not C: Gives the area of the triangle alone, forgetting to include the sector at all.
    • Why not D: Adds the triangle's area to the sector's area instead of subtracting it, rather than removing the triangle to leave just the segment.

  7. Question 7Answer: D

    1. From L2, y = x - 1. Substituting into L1: 2x + (x - 1) = 8, so 3x - 1 = 8, giving x = 3 and y = 2. So P = (3, 2).
    2. A line parallel to y = 3x - 5 has the same gradient, 3.
    3. Using P = (3, 2) with gradient 3: y - 2 = 3(x - 3), which expands to y - 2 = 3x - 9.
    4. So y = 3x - 7, option D.
    • Why not A: Rearranges L2 (x - y = 1) incorrectly as y = 1 - x instead of y = x - 1 when substituting into L1, a sign error that gives the wrong intersection point (7, -6) instead of (3, 2).
    • Why not B: Finds the correct intersection point (3, 2), but uses the perpendicular gradient -1/3 instead of maintaining the given parallel gradient of 3.
    • Why not C: Finds the correct point (3, 2) and the correct gradient 3, but expands y - 2 = 3(x - 3) as y - 2 = 3x - 3, forgetting to multiply the second term by 3 as well, giving y = 3x - 1 instead of y = 3x - 7.

  8. Question 8Answer: C

    1. Let c = cos(x). The equation becomes 2c^2 - c - 1 = 0, which factorises as (2c + 1)(c - 1) = 0.
    2. So cos(x) = -1/2 or cos(x) = 1.
    3. cos(x) = 1 gives x = 0 or x = 360 (the endpoints of the range).
    4. cos(x) = -1/2 has reference angle 60 degrees; cosine is negative in the second and third quadrants, giving x = 180 - 60 = 120 or x = 180 + 60 = 240.
    5. Combining all four: x = 0, 120, 240 or 360 degrees, option C.
    • Why not A: Correctly solves the factor 2cos(x) + 1 = 0, but discards the factor cos(x) - 1 = 0 as though it gave no valid solutions, missing x = 0 and x = 360.
    • Why not B: Correctly solves the factor cos(x) - 1 = 0, but discards the factor 2cos(x) + 1 = 0 as though it gave no valid solutions, missing x = 120 and x = 240.
    • Why not D: Correctly solves cos(x) = 1 for x = 0 and 360, but places the reference angle for cos(x) = -1/2 in the first and fourth quadrants, where cosine is positive, instead of the second and third, where it is negative, giving x = 60 and x = 300 instead of x = 120 and x = 240.

  9. Question 9Answer: B

    1. Factorise the numerator as a difference of two squares: 1 - sin^2(theta) = (1 - sin(theta))(1 + sin(theta)).
    2. So the expression becomes [(1 - sin(theta))(1 + sin(theta))] / (1 - sin(theta)).
    3. Since sin(theta) != 1, the factor (1 - sin(theta)) is nonzero and cancels, leaving 1 + sin(theta), option B.
    • Why not A: Treats 1 - sin^2(theta) as (1 - sin(theta))^2 rather than as the product of two different factors, (1 - sin(theta))(1 + sin(theta)), and so cancels to leave 1 - sin(theta) instead of 1 + sin(theta).
    • Why not C: Correctly factorises the numerator as (1 - sin(theta))(1 + sin(theta)), but cancels the wrong factor against the denominator, leaving (1 - sin(theta))(1 - sin(theta)) instead of (1 + sin(theta)).
    • Why not D: Uses sin^2(theta) + cos^2(theta) = 1 to rewrite the numerator as cos^2(theta), then takes this directly to cos(theta) and treats the (1 - sin(theta)) denominator as cancelling completely, rather than factorising the original numerator and cancelling properly.

  10. Question 10Answer: D

    1. y = tan(x) is undefined wherever cos(x) = 0, since tan(x) = sin(x)/cos(x).
    2. Within -180 <= x <= 180 degrees, cos(x) = 0 at x = -90 and x = 90.
    3. At the endpoints, cos(-180) = cos(180) = -1, so tan(x) = 0 there, not undefined.
    4. So the graph has exactly 2 vertical asymptotes in this interval, at x = -90 and x = 90 degrees, option D.
    • Why not A: Mistakenly believes tan(x) is undefined at the interval's endpoints -180 and 180 as well as at -90 and 90, whereas tan(-180) = tan(180) = 0, so the graph is perfectly well-defined there.
    • Why not B: Forgets that tan(x) has period 180 degrees, so an asymptote at x = 90 recurs every 180 degrees; this misses the equivalent asymptote at x = 90 - 180 = -90 within the given range.
    • Why not C: Confuses the interval's endpoints, where tan(x) = 0, with the values where the graph is actually undefined, which occur where cos(x) = 0.

  11. Question 11Answer: A

    1. The perpendicular from the centre of a circle to a chord bisects the chord, so half the chord length is 10/2 = 5 cm.
    2. This half-chord, the perpendicular distance (4 cm) and the radius form a right-angled triangle, with the radius as the hypotenuse.
    3. By Pythagoras' theorem: r^2 = 5^2 + 4^2 = 25 + 16 = 41.
    4. So r = sqrt(41) cm, option A.
    • Why not B: Forgets that the perpendicular from the centre bisects the chord, and so uses the full chord length 10 as one leg of the right-angled triangle instead of the half-chord length 5, giving r = sqrt(10^2 + 4^2) = sqrt(116) = 2sqrt(29).
    • Why not C: Subtracts instead of adds when combining the half-chord and the perpendicular distance, computing r^2 = 5^2 - 4^2 = 9, treating the radius as if it were a leg of the right triangle rather than its hypotenuse.
    • Why not D: Adds the half-chord length and the perpendicular distance directly instead of applying Pythagoras' theorem, giving r = 5 + 4 = 9 instead of r = sqrt(5^2 + 4^2).

  12. Question 12Answer: C

    1. Since angles A, B and C sum to 180 degrees, angle C = 180 - 30 - 90 = 60 degrees.
    2. By the sine rule, c/sin(C) = a/sin(A), so c = a sin(C)/sin(A) = 5 sin(60)/sin(30) = 5 x (sqrt(3)/2)/(1/2) = 5sqrt(3).
    3. Angle B lies between sides a and c, so the area is (1/2) a c sin(B) = (1/2)(5)(5sqrt(3)) sin(90) = (1/2)(25sqrt(3))(1).
    4. This gives an area of (25sqrt(3))/2 cm^2, option C.
    • Why not A: Forgets to find the second side c using the sine rule, and instead calculates the area using side a twice: (1/2)(5)(5) sin(90) = 25/2.
    • Why not B: Inverts the sine rule ratio when finding side c, computing c = a sin(A)/sin(C) instead of c = a sin(C)/sin(A), giving c = 5sqrt(3)/3 instead of 5sqrt(3), and using this smaller value in the area formula.
    • Why not D: Correctly finds c = 5sqrt(3) using the sine rule, but forgets the leading factor of 1/2 in the area formula (1/2) a c sin(B), giving (5)(5sqrt(3))(1) instead of (1/2)(5)(5sqrt(3))(1).

  13. Question 13Answer: D

    1. Substitute y = x - 1 into x^2 + y^2 = 25: x^2 + (x - 1)^2 = 25, which expands to 2x^2 - 2x - 24 = 0, or x^2 - x - 12 = 0.
    2. Factorising: (x - 4)(x + 3) = 0, so x = 4 or x = -3, giving points (4, 3) and (-3, -4).
    3. The distance between them is sqrt((4 - (-3))^2 + (3 - (-4))^2) = sqrt(7^2 + 7^2) = sqrt(98).
    4. sqrt(98) = sqrt(49 x 2) = 7sqrt(2), option D.
    • Why not A: Correctly finds the two intersection points (4, 3) and (-3, -4), but computes only the horizontal distance between them, 4 - (-3) = 7, forgetting to include the vertical difference in the distance formula.
    • Why not B: Correctly computes the sum of squares in the distance formula, 7^2 + 7^2 = 98, but forgets to take the square root, giving 98 instead of sqrt(98).
    • Why not C: Adds the horizontal and vertical differences between the two points directly, 7 + 7 = 14, instead of squaring each, summing, and then taking the square root.

  14. Question 14Answer: A

    1. sqrt(3)tan(x) = 3 gives tan(x) = 3/sqrt(3) = sqrt(3), after rationalising the denominator.
    2. The reference angle is 60 degrees, since tan(60) = sqrt(3).
    3. tan(x) is positive in the first and third quadrants. Within 0 <= x <= 360, the first-quadrant solution is x = 60 degrees, and the third-quadrant solution is x = 180 + 60 = 240 degrees.
    4. So the solutions are x = 60 degrees or x = 240 degrees, option A.
    • Why not B: Correctly finds the reference angle 60 degrees and the first solution x = 60, but forgets that tan(x) has period 180 degrees, so misses the second solution x = 180 + 60 = 240 within the given range.
    • Why not C: Confuses tan(60) = sqrt(3) with tan(30) = 1/sqrt(3), and so uses a reference angle of 30 degrees instead of 60, giving x = 30 or x = 210 instead of x = 60 or x = 240.
    • Why not D: Correctly finds the reference angle 60 degrees, but places the solutions in the quadrants where tangent is negative (the second and fourth) instead of where it is positive (the first and third), giving x = 120 and x = 300 instead of x = 60 and x = 240.

  15. Question 15Answer: C

    1. Since AB is a diameter, the centre of the circle is the midpoint of A and B: ((-4 + 6)/2, (1 + 1)/2) = (1, 1).
    2. The radius is half the length of AB: AB = sqrt((6 - (-4))^2 + (1 - 1)^2) = sqrt(100) = 10, so the radius is 5.
    3. The circle's equation is (x - 1)^2 + (y - 1)^2 = 25. Substituting y = 5: (x - 1)^2 + 16 = 25, so (x - 1)^2 = 9.
    4. Taking both square roots: x - 1 = 3 or x - 1 = -3, giving x = 4 or x = -2, option C.
    • Why not A: Computes the centre of the circle by adding the coordinates of A and B without dividing by 2, using (2, 2) as the centre instead of the correct midpoint (1, 1).
    • Why not B: Forgets to subtract the centre's y-coordinate before squaring, using y^2 = 25 in place of (y - 1)^2 = 16, which gives (x - 1)^2 = 0 and so x = 1 only.
    • Why not D: Correctly reaches (x - 1)^2 = 9, but takes only the positive square root, x - 1 = 3, giving x = 4 only and missing the second solution x = -2.

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