Sine rule, cosine rule and area of a triangle – Year 11 GCSE maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. \(\dfrac b{\sin70^\circ}=\dfrac{12}{\sin50^\circ}\) (1), \(b=\dfrac{12\sin70^\circ}{\sin50^\circ}\) (1) \(=\) 14.72 cm (1) We know \(a\) and its opposite angle \(A\), so the sine rule gives \(b\), the side opposite \(B\). Multiply both sides by \(\sin70^\circ\).
   2. \(C=180-50-70=\) \(60^\circ\) (1) The angles of a triangle add up to \(180^\circ\).

   Check (a): the side opposite the larger angle is longer, and \(70^\circ\gt 50^\circ\) gives \(b\gt a\). The side \(c\) is \(12\sin60^\circ\div\sin50^\circ\approx13.57\) cm.

2. Answer to Question 2 (4 marks)

   1. \(a^2=81+121-2\times9\times11\cos62^\circ\) (1) \(=202-92.96=109.04\) (1), \(a=\) 10.44 cm (1) Two sides and the included angle are known, so we use the cosine rule, then take the square root.
   2. \(\tfrac12\times9\times11\sin62^\circ=\) 43.7 cm² (1) Use \(\tfrac12bc\sin A\) with the two given sides and the angle between them.

   In (a), a common error is to work out \((81+121-198)\cos62^\circ\) by ignoring the order of operations. Keep the calculator value of \(a\) for later parts.

3. Answer to Question 3 (4 marks)

   1. \(\tfrac12\times7.5\times9\sin42^\circ=\) 22.6 cm² (2) Two sides and the included angle are known, so we use \(\tfrac12ab\sin C\).
   2. \(c^2=56.25+81-135\cos42^\circ=137.25-100.33=36.92\) (1), \(c=\) 6.08 cm (1) Use the cosine rule with the same two sides and angle, then take the square root.

   In (b), check that the answer is sensible: the third side is shorter than \(7.5+9=16.5\) cm and the angle opposite it, \(42^\circ\), is the smallest angle, so it should be the shortest side, which 6.08 cm is.

4. Answer to Question 4 (4 marks)

   1. \(\cos x=\dfrac{36+64-121}{2\times6\times8}\) (1) \(=\dfrac{-21}{96}\) (1), \(x=\) \(102.6^\circ\) (1) All three sides are known, so we use the rearranged cosine rule. The largest angle is opposite the longest side, 11 cm.
   2. Obtuse (1) The largest angle is more than \(90^\circ\) because its cosine is negative.

   In (b), the converse of Pythagoras gives the same result: \(6^2+8^2=100\), which is less than \(11^2=121\), so the angle is obtuse.

5. Answer to Question 5 (4 marks)

   1. \(d^2=225+400-600\cos75^\circ\) (1) \(=625-155.3=469.7\) (1), \(d=\) 21.7 km (1) The two paths and the angle between them are known, so we use the cosine rule.
   2. \(\tfrac12\times15\times20\sin75^\circ=\) 144.9 km² (1) Use \(\tfrac12ab\sin C\) with the two paths and the \(75^\circ\) angle.

   A common error in (a) is to use Pythagoras, which only works if the angle is \(90^\circ\). In (b), the area is in square kilometres because the lengths are in km.

Total : 20 marks