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ESAT Maths 2 Trigonometry questions

12 original ESAT Maths 2 trigonometry questions in the real exam style, each with a step-by-step worked solution and an explanation of why the tempting wrong answers are wrong. Try each one before opening the solution.

Methods on this page: Reduce to sin or cos = constant, then count per periodSubstitute to get a quadratic, then use the substitute's rangeUse the cosine rule and ½ab sin CArc length, sector and segment in radiansSine rule: the ambiguous case (two possible triangles)

Question 1ESAT Maths 2Warm-up

In triangle ABCABC, BC=9BC = 9, CA=15CA = 15 and angle ACB=120∘ACB = 120^\circ. What is the length ABAB?
  1. A171\sqrt{171}
  2. B306\sqrt{306}
  3. C1919
  4. D2020
  5. E2121
  6. F2222
  7. G2323
  8. H2424
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Answer: E · Method: Use the cosine rule and ½ab sin C

  1. Cosine rule: AB2=92+152−2×9×15cos⁡120∘AB^2 = 9^2 + 15^2 - 2 \times 9 \times 15 \cos 120^\circ.
  2. cos⁡120∘=−12\cos 120^\circ = -\tfrac{1}{2}, so AB2=306+135=441AB^2 = 306 + 135 = 441 and AB=21AB = 21.

Why the tempting wrong answers are wrong

  • A: cos⁡120∘=−12\cos 120^\circ = -\tfrac{1}{2}. Check the sign in the cosine rule.
  • B: That's Pythagoras, which only works for a right angle.

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Question 2ESAT Maths 2Warm-up

A sector of a circle has radius 66 and angle 2π3\dfrac{2\pi}{3} radians. What is its perimeter?
  1. A4π4\pi
  2. B4π+64\pi + 6
  3. C4π+124\pi + 12
  4. D8π+248\pi + 24
  5. E12π+1212\pi + 12
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Answer: C · Method: Arc length, sector and segment in radians

  1. Arc length =rθ=6×2π3=4π= r\theta = 6 \times \dfrac{2\pi}{3} = 4\pi.
  2. Perimeter == arc +2r=4π+12+ 2r = 4\pi + 12.

Why the tempting wrong answers are wrong

  • A: That's just the arc. The perimeter also includes the two radii.
  • B: There are two straight edges, each a radius.

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Question 3ESAT Maths 2Warm-up

In triangle ABCABC, BC=9BC = 9, CA=24CA = 24 and angle ACB=60∘ACB = 60^\circ. What is the area of the triangle?
  1. A27327\sqrt{3}
  2. B5454
  3. C54354\sqrt{3}
  4. D108108
  5. E1083108\sqrt{3}
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Answer: C · Method: Use the cosine rule and ½ab sin C

  1. Area =12×9×24×sin⁡60∘= \tfrac{1}{2} \times 9 \times 24 \times \sin 60^\circ, and sin⁡60∘=32\sin 60^\circ = \tfrac{\sqrt{3}}{2}.
  2. That's 54354\sqrt{3}.

Why the tempting wrong answers are wrong

  • B: sin⁡60∘=32\sin 60^\circ = \tfrac{\sqrt{3}}{2}, not 12\tfrac{1}{2}.
  • E: The formula is 12absin⁡C\tfrac{1}{2}ab\sin C: don't forget the 12\tfrac{1}{2}.

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Question 4ESAT Maths 2Standard

How many solutions does 2cos⁡2x−5cos⁡x+2=02\cos^2 x - 5\cos x + 2 = 0 have for 0≤x≤4π0 \le x \le 4\pi?
  1. A00
  2. B22
  3. C33
  4. D44
  5. E55
  6. F66
  7. G77
  8. H88
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Answer: D · Method: Reduce to sin or cos = constant, then count per period

  1. Factorise: cos⁡x=2\cos x = 2 or cos⁡x=12\cos x = \dfrac{1}{2}.
  2. cos⁡x=2\cos x = 2 is impossible. cos⁡x=12\cos x = \dfrac{1}{2} has 4 solutions in the interval.
  3. That's 44 solutions in total.

Why the tempting wrong answers are wrong

  • B: The interval is 4π4\pi long, not one period of 2π2\pi.

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Question 5ESAT Maths 2Standard

As xx varies over the real numbers, what is the greatest value of cos⁡2x+2cos⁡x+1\cos^2 x + 2\cos x + 1?
  1. A00
  2. B33
  3. C44
  4. D55
  5. E66
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Answer: C · Method: Substitute to get a quadratic, then use the substitute's range

  1. Let s=cos⁡xs = \cos x, so the expression is s2+2s+1s^2 + 2s + 1.
  2. ss takes every value in −1≤s≤1-1 \le s \le 1. The turning point is at s=−1s = -1, which is inside this range.
  3. Check the turning point and the ends: the values are 44, 00, 00. The greatest is 44.

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Question 6ESAT Maths 2Standard

A circle has radius 66. A chord cuts off a minor segment, and the angle the chord subtends at the centre is π6\dfrac{\pi}{6} radians. What is the area of the minor segment?
  1. A3π2−92\dfrac{3\pi}{2} - \dfrac{9}{2}
  2. B3π−93\pi - 9
  3. C6π−186\pi - 18
  4. D99
  5. E3π3\pi
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Answer: B · Method: Arc length, sector and segment in radians

  1. Sector area =12r2θ=12×36×π6=3π= \tfrac{1}{2}r^2\theta = \tfrac{1}{2} \times 36 \times \dfrac{\pi}{6} = 3\pi.
  2. Triangle area =12r2sin⁡θ=12×36×12=9= \tfrac{1}{2}r^2\sin\theta = \tfrac{1}{2} \times 36 \times \tfrac{1}{2} = 9.
  3. Segment == sector −- triangle =3π−9= 3\pi - 9.

Why the tempting wrong answers are wrong

  • C: The sector is 12r2θ\tfrac{1}{2}r^2\theta and the triangle is 12r2sin⁡θ\tfrac{1}{2}r^2\sin\theta: don't forget the halves.
  • E: That's the whole sector. Subtract the triangle formed by the two radii and the chord.

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Question 7ESAT Maths 2Standard

How many solutions does 2cos⁡2x−5cos⁡x+2=02\cos^2 x - 5\cos x + 2 = 0 have for −π≤x≤π-\pi \le x \le \pi?
  1. A00
  2. B11
  3. C22
  4. D33
  5. E44
  6. F55
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Answer: C · Method: Reduce to sin or cos = constant, then count per period

  1. Factorise: cos⁡x=2\cos x = 2 or cos⁡x=12\cos x = \dfrac{1}{2}.
  2. cos⁡x=2\cos x = 2 is impossible. cos⁡x=12\cos x = \dfrac{1}{2} has 2 solutions in the interval.
  3. That's 22 solutions in total.

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Question 8ESAT Maths 2Standard

A circle has radius 33. A chord cuts off a minor segment, and the angle the chord subtends at the centre is 2π3\dfrac{2\pi}{3} radians. What is the area of the minor segment?
  1. A3π2−938\dfrac{3\pi}{2} - \dfrac{9\sqrt{3}}{8}
  2. B934\dfrac{9\sqrt{3}}{4}
  3. C3π−9343\pi - \dfrac{9\sqrt{3}}{4}
  4. D3π3\pi
  5. E6π−9326\pi - \dfrac{9\sqrt{3}}{2}
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Answer: C · Method: Arc length, sector and segment in radians

  1. Sector area =12r2θ=12×9×2π3=3π= \tfrac{1}{2}r^2\theta = \tfrac{1}{2} \times 9 \times \dfrac{2\pi}{3} = 3\pi.
  2. Triangle area =12r2sin⁡θ=12×9×32=934= \tfrac{1}{2}r^2\sin\theta = \tfrac{1}{2} \times 9 \times \tfrac{\sqrt{3}}{2} = \dfrac{9\sqrt{3}}{4}.
  3. Segment == sector −- triangle =3π−934= 3\pi - \dfrac{9\sqrt{3}}{4}.

Why the tempting wrong answers are wrong

  • D: That's the whole sector. Subtract the triangle formed by the two radii and the chord.
  • E: The sector is 12r2θ\tfrac{1}{2}r^2\theta and the triangle is 12r2sin⁡θ\tfrac{1}{2}r^2\sin\theta: don't forget the halves.

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Question 9ESAT Maths 2Challenging

As xx varies over the real numbers, what is the difference between the greatest and least values of 2sin⁡2x+3cos⁡x−42\sin^2 x + 3\cos x - 4?
  1. A−78-\dfrac{7}{8}
  2. B418\dfrac{41}{8}
  3. C66
  4. D498\dfrac{49}{8}
  5. E578\dfrac{57}{8}
  6. F658\dfrac{65}{8}
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Answer: D · Method: Substitute to get a quadratic, then use the substitute's range

  1. Use sin⁡2x=1−cos⁡2x\sin^2 x = 1 - \cos^2 x: the expression is −2s2+3s−2-2s^2 + 3s - 2 with s=cos⁡xs = \cos x.
  2. ss takes every value in −1≤s≤1-1 \le s \le 1. The turning point is at s=34s = \dfrac{3}{4}, which is inside this range.
  3. Check the turning point and the ends: the values are −1-1, −7-7, −78-\dfrac{7}{8}. The greatest is −78-\dfrac{7}{8} and the least is −7-7, a difference of 498\dfrac{49}{8}.

Why the tempting wrong answers are wrong

  • C: Check the turning point as well as the ends: it's inside the range.

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Question 10ESAT Maths 2Challenging

In triangle ABCABC, angle BAC=45∘BAC = 45^\circ, AB=12AB = 12 and BC=9BC = 9. How many different triangles fit this description?
  1. A00
  2. B11
  3. C22
  4. D33
  5. E44
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Answer: C · Method: Sine rule: the ambiguous case (two possible triangles)

  1. The shortest distance from BB to the line ACAC is 12sin⁡45∘=6212\sin 45^\circ = 6\sqrt{2}.
  2. BC=9BC = 9 is between that height and AB=12AB = 12, so CC can be in two places: two triangles.

Why the tempting wrong answers are wrong

  • B: This is the ambiguous case: BCBC is shorter than ABAB but longer than the height from BB, so it can swing to two positions.

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Question 11ESAT Maths 2Challenging

In triangle ABCABC, angle BAC=30∘BAC = 30^\circ, AB=4AB = 4 and BC=3BC = 3. There are two possible lengths for ACAC. What is their sum?
  1. A423\tfrac{4}{2}\sqrt{3}
  2. B44
  3. C66
  4. D434\sqrt{3}
  5. E838\sqrt{3}
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Answer: D · Method: Sine rule: the ambiguous case (two possible triangles)

  1. Cosine rule with AC=xAC = x: 32=x2+42−2×4xcos⁡30∘3^2 = x^2 + 4^2 - 2 \times 4 x \cos 30^\circ, a quadratic in xx.
  2. Its two roots are the two possible lengths, and their sum is 2×4cos⁡30∘=432 \times 4 \cos 30^\circ = 4\sqrt{3}.

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Question 12ESAT Maths 2Challenging

As xx varies over the real numbers, what is the greatest value of 2cos⁡2x−6sin⁡x−62\cos^2 x - 6\sin x - 6?
  1. A−12-12
  2. B−2-2
  3. C−1-1
  4. D00
  5. E12\dfrac{1}{2}
  6. F11
  7. G22
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Answer: D · Method: Substitute to get a quadratic, then use the substitute's range

  1. Use cos⁡2x=1−sin⁡2x\cos^2 x = 1 - \sin^2 x: the expression is −2s2−6s−4-2s^2 - 6s - 4 with s=sin⁡xs = \sin x.
  2. ss takes every value in −1≤s≤1-1 \le s \le 1. The turning point is at s=−32s = -\dfrac{3}{2}, which is outside this range.
  3. So the extremes are at the ends: s=1s = 1 gives −12-12 and s=−1s = -1 gives 00. The greatest is 00.

Why the tempting wrong answers are wrong

  • E: The turning point is at sin⁡x=−32\sin x = -\dfrac{3}{2}, which sin⁡x\sin x can never reach.

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