Trigonometry – Year 12 A level maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. Two sides and the included angle are known, so use the cosine rule. \(BC^2=49+100-140\cos58^\circ=149-74.19=74.81\) (1), \(BC=8.65\) cm (1) So \(BC\approx8.65\) cm.
   2. Use area \(=\tfrac12bc\sin A\) with the two given sides and the angle between them. Area \(=\tfrac12\times7\times10\sin58^\circ\) (1) \(=29.7\) cm\(^2\) (1) The area is about 29.7 cm\(^2\).

   Make sure the calculator is in degree mode; in radian mode \(\cos58\) gives a nonsense length. Keep \(BC^2\) unrounded before taking the square root.

2. Answer to Question 2 (4 marks)

   1. In radians, arc length \(=r\theta\). \(9\times0.8=7.2\) cm (1) The arc is 7.2 cm long.
   2. In radians, sector area \(=\tfrac12r^2\theta\). \(\tfrac12\times81\times0.8=32.4\) cm\(^2\) (1) The sector has area 32.4 cm\(^2\).
   3. Segment \(=\) sector \(-\) triangle \(=\tfrac12r^2(\theta-\sin\theta)\), with the calculator in radians. Segment \(=\tfrac12\times81(0.8-\sin0.8)\) (1) \(=40.5\times0.0827=3.35\) cm\(^2\) (1) The segment has area about 3.35 cm\(^2\).

   Check: the triangle has area \(\tfrac12\times81\sin0.8\approx29.05\) cm\(^2\), and \(32.4-29.05=3.35\) ✓.

3. Answer to Question 3 (4 marks)

   Start with the left-hand side and add the fractions over the common denominator \(\sin\theta(1+\cos\theta)\); then use \(\sin^2\theta+\cos^2\theta=1\).

   LHS \(=\dfrac{\sin^2\theta+(1+\cos\theta)^2}{\sin\theta(1+\cos\theta)}\) (1) \(=\dfrac{\sin^2\theta+1+2\cos\theta+\cos^2\theta}{\sin\theta(1+\cos\theta)}\) (1) \(=\dfrac{2+2\cos\theta}{\sin\theta(1+\cos\theta)}\) (1) \(=\dfrac{2(1+\cos\theta)}{\sin\theta(1+\cos\theta)}=\dfrac2{\sin\theta}\) (1).

   The factor \((1+\cos\theta)\) cancels, which gives the right-hand side and proves the identity.

   The working should go from one side to the other, not change both sides at once. A useful check: at \(\theta=90^\circ\) the left-hand side is \(1+1=2\) and the right-hand side is \(\tfrac21=2\) ✓.

4. Answer to Question 4 (4 marks)

   Use \(\sin^2\theta=1-\cos^2\theta\) to obtain a quadratic in \(\cos\theta\), then factorise.

   \(2(1-\cos^2\theta)=1+\cos\theta\) (1), so \(2\cos^2\theta+\cos\theta-1=0\) (1), \((2\cos\theta-1)(\cos\theta+1)=0\) (1). \(\cos\theta=\tfrac12\): \(60^\circ,300^\circ\); \(\cos\theta=-1\): \(180^\circ\) (1).

   So the solutions are \(\theta=60^\circ\), \(180^\circ\) and \(300^\circ\). Check \(180^\circ\): \(2\times0=1+(-1)\) ✓.

   Note that \(\cos\theta=-1\) has only one solution in this interval.

   Dividing both sides by \(\cos\theta+1\) at the start would lose the solution \(180^\circ\). Check \(60^\circ\): \(2\times\tfrac34=1.5=1+\tfrac12\) ✓.

5. Answer to Question 5 (4 marks)

   1. The bearing of \(H\) from \(A\) is the back-bearing \(065^\circ+180^\circ=245^\circ\); the boat then heads on \(140^\circ\). Back bearing at \(A\) is \(245^\circ\); \(245^\circ-140^\circ=105^\circ\) (1) So angle \(HAB=105^\circ\).
   2. Use the cosine rule with \(HA=12\), \(AB=9\) and the included angle \(105^\circ\). \(HB^2=144+81-2\times12\times9\cos105^\circ=225+55.9=280.9\) (1), \(HB=16.8\) km (1) So \(HB\approx16.8\) km.
   3. Find angle \(AHB\) with the sine rule, then add it to the bearing of \(A\) from \(H\). \(\sin H=\dfrac{9\sin105^\circ}{16.76}=0.5187\), \(H=31.2^\circ\), bearing \(065^\circ+31^\circ=096^\circ\) (1) The bearing of \(B\) from \(H\) is \(096^\circ\).

Total : 20 marks