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

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1. Answer to Question 1

   1. \(\dfrac b{\sin65^\circ}=\dfrac8{\sin40^\circ}\), so \(b=\dfrac{8\sin65^\circ}{\sin40^\circ}=\) 11.28 cm. We know a side and its opposite angle (\(a\) and \(A\)) and want the side opposite \(B\), so the sine rule links them. Multiply both sides by \(\sin65^\circ\).
   2. \(C=180-40-65=\) \(75^\circ\). The angles of a triangle add up to \(180^\circ\).
   3. \(c=\dfrac{8\sin75^\circ}{\sin40^\circ}=\) 12.02 cm. Use the sine rule again with the pair \(a\) and \(A\), which are exact given values, rather than the rounded \(b\).

2. Answer to Question 2

   1. \(\dfrac{\sin B}{12}=\dfrac{\sin35^\circ}9\), so \(\sin B=\dfrac{12\sin35^\circ}9=0.7648\) and \(B=\) \(49.9^\circ\). We know \(a\) and its opposite angle \(A\), and the side \(b\), so the sine rule gives \(\sin B\). Use \(\sin^{-1}\) for the acute value.
   2. \(\sin B=0.7648\) also for \(180-49.9=\) \(130.1^\circ\), because \(\sin x=\sin(180-x)\). Both give valid triangles here since \(35+130.1\lt 180\). This is the ambiguous case of the sine rule: two different triangles fit the same information.
   3. The acute value, \(49.9^\circ\). So when the triangle is acute, angle \(B=49.9^\circ\) and angle \(C=180-35-49.9=95.1^\circ\) would not be acute, which shows we must check the other angles too.

3. Answer to Question 3

   1. \(a^2=49+81-2\times7\times9\cos48^\circ=130-84.31=45.69\), so \(a=\) 6.76 cm. We know two sides and the angle between them, so we use the cosine rule.
   2. \(\sin B=\dfrac{7\sin48^\circ}{6.76}=0.7696\), so \(B=\) \(50.3^\circ\). Now we know a side and its opposite angle, so we can use the sine rule. \(B\) is not the largest angle, so it is acute.
   3. \(C=180-48-50.3=81.7^\circ\). Check with the sine rule: \(\sin C=\dfrac{9\sin48^\circ}{6.76}=0.989\), so \(C=81.7^\circ\) ✓. The angles of a triangle add up to \(180^\circ\).

4. Answer to Question 4

   1. \(\cos C=\dfrac{a^2+b^2-c^2}{2ab}\). Rearrange \(c^2=a^2+b^2-2ab\cos C\) to make \(\cos C\) the subject.
   2. \(\cos C=\dfrac{49+64-100}{2\times7\times8}=\dfrac{13}{112}\), so \(C=\) \(83.3^\circ\). The largest angle is opposite the longest side, 10 cm. Use \(\cos^{-1}\) on the calculator.
   3. \(\cos A=\dfrac{64+100-49}{2\times8\times10}=0.71875\), so \(A=\) \(44.0^\circ\). \(B=180-83.3-44.0=\) \(52.7^\circ\). Use the cosine rule again for \(A\), then subtract from \(180^\circ\) to find \(B\).

5. Answer to Question 5

   1. \(\tfrac12\times8\times11\times\sin37^\circ=\) 26.5 cm². The two sides and the included angle are known, so we can use \(\tfrac12ab\sin C\) directly.
   2. \(12=\tfrac12\times5\times6\sin C=15\sin C\), so \(\sin C=0.8\) and \(C=\) \(53.1^\circ\). Substitute the area and the sides, then divide by 15 and use \(\sin^{-1}\).
   3. \(\sin C\le1\), so the area is at most \(\tfrac12\times5\times6=15\) cm². The largest possible value of sine is 1.

6. Answer to Question 6

   1. \(\tfrac12\times4\times6\times\sin30^\circ=12\times\dfrac12=\) 6 cm². Use \(\sin30^\circ=\tfrac12\) from the table of exact values.
   2. \(12\sin60^\circ=12\times\dfrac{\sqrt3}2=\) \(6\sqrt3\) cm² \(=10.39\). Use \(\sin60^\circ=\tfrac{\sqrt3}{2}\).
   3. \(12\times1=\) 12 cm². The largest area is when \(\sin C=1\), that is \(C=90^\circ\). \(\sin90^\circ=1\), the largest value sine can take.

7. Answer to Question 7

   1. \(110-50=\) \(60^\circ\). Both bearings are measured from north at \(P\), so the angle between the two paths is the difference of the bearings.
   2. \(AB^2=12^2+9^2-2\times12\times9\cos60^\circ=225-108=117\), so \(AB=\) 10.82 km. Two sides and the included angle are known, so we use the cosine rule; \(\cos60^\circ=0.5\).
   3. \(\tfrac12\times12\times9\sin60^\circ=\) 46.8 km². Use \(\tfrac12ab\sin C\) with the two paths and the angle between them.
   4. \(\sin A=\dfrac{9\sin60^\circ}{10.82}=0.7205\), so \(A=\) \(46.1^\circ\). Use the sine rule with the side \(AB\) and its opposite angle \(60^\circ\).

8. Answer to Question 8

   1. \(\tfrac12\times120\times150\sin52^\circ=\) 7 092 m². Two sides and the included angle are known, so we use \(\tfrac12ab\sin C\).
   2. \(c^2=120^2+150^2-2\times120\times150\cos52^\circ=36\,900-22\,164=14\,736\), \(c=\) 121.4 m. Two sides and the included angle are known, so we use the cosine rule for the third side.
   3. Perimeter \(=120+150+121.4=391.4\) m. Cost \(=391.4\times12.50=\) £4 892.50 (about). Add the three sides for the perimeter, then multiply by the price per metre.

9. Answer to Question 9

   1. \(180-62-71=\) \(47^\circ\). The angles of a triangle add up to \(180^\circ\).
   2. \(\dfrac{AT}{\sin71^\circ}=\dfrac{200}{\sin47^\circ}\), so \(AT=\) 258.6 m. \(AT\) is opposite the \(71^\circ\) angle at \(B\), and \(AB=200\) m is opposite the \(47^\circ\) angle.
   3. \(BT=\dfrac{200\sin62^\circ}{\sin47^\circ}=\) 241.5 m. \(BT\) is opposite the \(62^\circ\) angle at \(A\).
   4. Height \(=258.6\tan12^\circ=\) 55.0 m. The height is opposite the \(12^\circ\) angle and \(AT\) is adjacent in the vertical right-angled triangle, so we use tangent.

10. Answer to Question 10

    1. The largest angle is opposite 5.0: \(\cos x=\dfrac{3.2^2+4.1^2-5^2}{2\times3.2\times4.1}=\dfrac{2.05}{26.24}\), so \(x=\) \(85.5^\circ\). All three sides are known, so we use the cosine rule in the form for \(\cos x\).
    2. \(\tfrac12\times3.2\times4.1\times\sin85.5^\circ=\) 6.54 m². Use \(\tfrac12ab\sin C\) with the two sides next to the angle found in part (a).
    3. \(6\times6.54\times18=\) £706 (to the nearest pound). Multiply the area of one sail by 6, then by the price per square metre.

11. Answer to Question 11

    1. At the turn the back bearing is \(030+180=210^\circ\) and the new direction is \(100^\circ\). The angle between them is \(210-100=\) \(110^\circ\).
    2. \(d^2=250^2+180^2-2\times250\times180\cos110^\circ=62\,500+32\,400+30\,782=125\,682\), so \(d=\) 354.5 km.
    3. \(\sin x=\dfrac{180\sin110^\circ}{354.5}=0.4771\), so \(x=28.5^\circ\) at the start. The bearing is \(030+28.5=\) \(059^\circ\) (to the nearest degree, \(58.5^\circ\) rounds to \(059^\circ\)).

12. Answer to Question 12

    1. \(AB^2=85^2+120^2-2\times85\times120\cos74^\circ=21\,625-5\,623=16\,002\), \(AB=\) 126.5 m. Two sides and the included angle are known, so we use the cosine rule.
    2. \(\sin A=\dfrac{120\sin74^\circ}{126.5}=0.9119\). Use the cosine rule to avoid any doubt about acute or obtuse: \(\cos A=\dfrac{85^2+126.5^2-120^2}{2\times85\times126.5}=0.4106\), so \(A=\) \(65.8^\circ\). The cosine rule gives a unique angle, because cosine is negative for obtuse angles.
    3. \(\tfrac12\times85\times120\sin74^\circ=\) 4 902 m². Use \(\tfrac12ab\sin C\) with the two measured sides and the angle between them.

13. Answer to Question 13

    1. \(\cos C=\dfrac{25+49-100}{2\times5\times7}=\dfrac{-26}{70}\), so \(C=\) \(111.8^\circ\). All three sides are known, so we use the rearranged cosine rule. The largest angle is opposite the longest side, 10 cm.
    2. \(\cos C\lt 0\) only for angles between \(90^\circ\) and \(180^\circ\). So a negative cosine always means the angle is obtuse.
    3. \(\cos A=\dfrac{49+100-25}{2\times7\times10}=0.8857\), \(A=27.7^\circ\). \(B=180-111.8-27.7=\) \(40.5^\circ\). Find \(A\) with the cosine rule, then subtract both known angles from \(180^\circ\).

14. Answer to Question 14

    1. \(30=\tfrac12\times10\times8\times\sin C=40\sin C\), so \(\sin C=0.75\). Substitute the area and the two sides into \(\tfrac12ab\sin C\), then divide by 40.
    2. \(C=48.6^\circ\) or \(C=180-48.6=131.4^\circ\). Sine is positive for both acute and obtuse angles, and \(\sin C=\sin(180^\circ-C)\).
    3. Case 1: \(c^2=100+64-160\cos48.6^\circ=164-105.8=58.2\), \(c=7.63\) cm. Case 2: \(c^2=164-160\cos131.4^\circ=164+105.8=269.8\), \(c=16.4\) cm. Use the cosine rule with each angle. For the obtuse angle \(\cos C\) is negative, so the third side is longer.

15. Answer to Question 15

    1. \(AC^2=36+49-84\cos80^\circ=85-14.59=70.41\), so \(AC=\) 8.39 cm. In triangle \(ABC\) we know two sides and the included angle, so we use the cosine rule.
    2. \(\cos ACD=\dfrac{8.39^2+5^2-9^2}{2\times8.39\times5}=\dfrac{70.41+25-81}{83.9}=0.1717\), so \(ACD=\) \(80.1^\circ\). In triangle \(ACD\) all three sides are now known, so we use the rearranged cosine rule.
    3. Triangle \(ABC\): \(\tfrac12\times6\times7\sin80^\circ=20.68\). Triangle \(ACD\): \(\tfrac12\times8.39\times5\sin80.1^\circ=20.67\). Total \(=\) 41.3 cm². Split the quadrilateral along the diagonal \(AC\) and add the areas of the two triangles, each found with \(\tfrac12ab\sin C\).

16. Answer to Question 16

    1. \(b=\dfrac{6\sin60^\circ}{\sin45^\circ}=6\times\dfrac{\sqrt3/2}{\sqrt2/2}=6\times\dfrac{\sqrt3}{\sqrt2}=6\times\dfrac{\sqrt6}2=\) \(3\sqrt6\). Use the exact values \(\sin60^\circ=\tfrac{\sqrt3}2\) and \(\sin45^\circ=\tfrac{\sqrt2}2\), and \(\tfrac{\sqrt3}{\sqrt2}=\tfrac{\sqrt6}{2}\).
    2. \(C=75^\circ\). Area \(=\tfrac12\times6\times3\sqrt6\times\sin75^\circ=9\sqrt6\sin75^\circ\). The angles add up to \(180^\circ\), so \(C=180-45-60\). The two sides \(a\) and \(b\) enclose angle \(C\).
    3. \(3\sqrt6=\) 7.35 cm. Using \(\sqrt6\approx2.449\), \(3\sqrt6\approx7.35\).

17. Answer to Question 17

    1. \(a^2=25+64-2\times5\times8\cos120^\circ=89+40=129\) (since \(\cos120^\circ=-0.5\)). Two sides and the included angle are known, so we use the cosine rule. Subtracting a negative adds 40.
    2. \(\sqrt{129}=\) 11.36 cm. Take the square root of 129.
    3. \(\tfrac12\times5\times8\times\sin120^\circ=20\times\dfrac{\sqrt3}2=\) \(10\sqrt3\) cm². Use \(\tfrac12ab\sin C\) with the exact value \(\sin120^\circ=\tfrac{\sqrt3}{2}\).
    4. The smallest angle is opposite the 5 cm side: \(\sin x=\dfrac{5\sin120^\circ}{11.36}=0.3812\), so \(x=\) \(22.4^\circ\). The smallest angle is opposite the shortest side. It must be acute, because the triangle already has an obtuse angle.

18. Answer to Question 18

    1. \(\cos x=\dfrac{49+64-169}{2\times7\times8}=\dfrac{-56}{112}=-\dfrac12\), so \(x=120^\circ\). The largest angle is opposite the longest side, 13 cm. A cosine of \(-\tfrac12\) gives exactly \(120^\circ\).
    2. \(\tfrac12\times7\times8\sin120^\circ=28\times\dfrac{\sqrt3}2=\) \(14\sqrt3\) cm² \(=24.25\). Use \(\tfrac12ab\sin C\) with the sides 7 and 8 that enclose the \(120^\circ\) angle, and \(\sin120^\circ=\tfrac{\sqrt3}{2}\).
    3. \(\tfrac12\times13\times h=24.25\), so \(h=\) 3.73 cm. Use area \(=\tfrac12\times\) base \(\times\) height with the 13 cm side as the base.

19. Answer to Question 19

    1. Angle \(SAB=90-30=60^\circ\). At \(B\) the direction to \(A\) is \(270^\circ\) and to \(S\) is \(310^\circ\), so angle \(SBA=310-270=40^\circ\). Bearings are measured clockwise from north, and \(AB\) points due east (\(090^\circ\)), so angle \(SAB=90^\circ-30^\circ\).
    2. \(180-60-40=\) \(80^\circ\). The angles of a triangle add up to \(180^\circ\).
    3. \(AS=\dfrac{10\sin40^\circ}{\sin80^\circ}=\) 6.53 km. \(AS\) is opposite the \(40^\circ\) angle at \(B\), and \(AB=10\) km is opposite the \(80^\circ\) angle at \(S\).
    4. \(6.53\sin60^\circ=\) 5.65 km. The distance north is the side opposite the \(60^\circ\) angle in the right-angled triangle with hypotenuse \(AS\).

20. Answer to Question 20

    1. In the right-angled triangle \(BCD\), \(\sin B=\dfrac{CD}{a}\), so \(CD=a\sin B\).
    2. Area \(=\tfrac12\times AB\times CD=\tfrac12\times c\times a\sin B=\tfrac12ac\sin B\).
    3. \(\tfrac12ac\sin B=\tfrac12bc\sin A\). Divide by \(\tfrac12c\): \(a\sin B=b\sin A\). Divide by \(\sin A\sin B\): \(\dfrac a{\sin A}=\dfrac b{\sin B}\). This is the sine rule.