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

← Back to the worksheet

1. Answer to Question 1

   1. The angles of a triangle add up to \(180^\circ\). \(C=180^\circ-48^\circ-65^\circ=\) \(67^\circ\).
   2. Side \(b\) is opposite angle \(B\), and we know the pair \(a\) and \(A\); use \(\dfrac a{\sin A}=\dfrac b{\sin B}\). \(\dfrac b{\sin65^\circ}=\dfrac{12}{\sin48^\circ}\), so \(b=\dfrac{12\sin65^\circ}{\sin48^\circ}=\) 14.6 cm. So \(b\approx14.6\) cm.
   3. Use the sine rule again with the pair \(c\) and \(C=67^\circ\). \(c=\dfrac{12\sin67^\circ}{\sin48^\circ}=\) 14.9 cm. Check: the longest side \(c\) is opposite the largest angle \(C\) ✓.

2. Answer to Question 2

   1. Two sides and the included angle are known, so use \(c^2=a^2+b^2-2ab\cos C\). \(c^2=7^2+9^2-2\times7\times9\cos52^\circ=130-126\cos52^\circ=52.43\), so \(c=\) 7.24 cm.
   2. Three sides are known, so rearrange the cosine rule to find an angle. The smallest angle is opposite the shortest side (8 cm): \(\cos A=\dfrac{11^2+14^2-8^2}{2\times11\times14}=\dfrac{253}{308}\), so \(A=\) 34.8\(^\circ\).
   3. Use area \(=\tfrac12ab\sin C\) with the included angle. Area \(=\tfrac12\times6.4\times9.1\times\sin38^\circ=\) 17.9 cm\(^2\).
   4. Since \(\sin(180^\circ-C)=\sin C\), there are two angles with the same sine. \(\tfrac12\times8\times10\sin C=20\), so \(\sin C=0.5\) and \(C=\) \(30^\circ\) or \(150^\circ\). Both triangles have area 20 cm\(^2\).

3. Answer to Question 3

   1. Multiply degrees by \(\dfrac\pi{180}\) to get radians, and multiply radians by \(\dfrac{180}\pi\) to get degrees. \(135^\circ=135\times\dfrac\pi{180}=\) \(\dfrac{3\pi}4\). \(\dfrac{5\pi}6=\dfrac{5\times180^\circ}6=\) \(150^\circ\).
   2. In radians, arc length \(=r\theta\) and sector area \(=\tfrac12r^2\theta\). Arc \(=r\theta=8\times1.2=\) 9.6 cm. Area \(=\tfrac12r^2\theta=\tfrac12\times64\times1.2=\) 38.4 cm\(^2\).
   3. Rearrange \(s=r\theta\) to find \(\theta\), then use the area formula. \(\theta=\dfrac sr=\dfrac{15}6=\) 2.5 rad. Area \(=\tfrac12\times36\times2.5=\) 45 cm\(^2\).
   4. Segment area \(=\) sector area \(-\) triangle area \(=\tfrac12r^2\theta-\tfrac12r^2\sin\theta\); the calculator must be in radians. Segment \(=\tfrac12r^2(\theta-\sin\theta)=50(1.2-\sin1.2)=50(1.2-0.9320)=\) 13.40 cm\(^2\). The segment has area 13.40 cm\(^2\).

4. Answer to Question 4

   1. Use the exact values \(\sin60^\circ=\cos30^\circ=\dfrac{\sqrt3}2\) and \(\tan45^\circ=1\). \(\dfrac{\sqrt3}2\times\dfrac{\sqrt3}2-1=\dfrac34-1=\) \(-\dfrac14\). So the exact value is \(-\tfrac14\).
   2. Use \(\tan\tfrac\pi3=\sqrt3\) and \(\sin\tfrac\pi4=\dfrac{\sqrt2}2\); dividing by a fraction means multiplying by its reciprocal. \(\dfrac{\sqrt3}{\sqrt2/2}=\dfrac{2\sqrt3}{\sqrt2}=\sqrt2\times\sqrt3=\) \(\sqrt6\). Note that \(\dfrac2{\sqrt2}=\sqrt2\).
   3. Use the related acute angle and the CAST diagram: \(150^\circ\) is in the second quadrant, where only sine is positive; \(210^\circ\) is in the third quadrant, where only tangent is positive; \(135^\circ\) is in the second quadrant. \(\cos150^\circ=-\cos30^\circ=\) \(-\dfrac{\sqrt3}2\). \(\sin210^\circ=-\sin30^\circ=\) \(-\dfrac12\). \(\tan135^\circ=-\tan45^\circ=\) \(-1\). These signs agree with the CAST diagram.

5. Answer to Question 5

   1. Use \(\sin^2\theta+\cos^2\theta\equiv1\) to replace the numerator, then cancel \(\sin\theta\). LHS \(=\dfrac{\sin^2\theta}{\sin\theta\cos\theta}=\dfrac{\sin\theta}{\cos\theta}=\tan\theta=\) RHS. This proves the identity.
   2. Write \(\tan\theta\) as \(\dfrac{\sin\theta}{\cos\theta}\), then add the fractions over a common denominator. LHS \(=\dfrac{\sin\theta}{\cos\theta}+\dfrac{\cos\theta}{\sin\theta}=\dfrac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}=\dfrac1{\sin\theta\cos\theta}=\) RHS. This proves the identity.
   3. Use \(\cos^2\theta=1-\sin^2\theta\); an acute angle has a positive cosine. \(\cos^2\theta=1-\dfrac9{25}=\dfrac{16}{25}\), and \(\theta\) is acute so \(\cos\theta=\) \(\dfrac45\). \(\tan\theta=\dfrac{3/5}{4/5}=\) \(\dfrac34\). This is the 3-4-5 right-angled triangle.

6. Answer to Question 6

   1. Use \(\sin^{-1}\) for the principal value; sine is also positive in the second quadrant, giving \(180^\circ-\alpha\). Principal value \(36.9^\circ\); the second solution is \(180^\circ-36.9^\circ=143.1^\circ\). So \(\theta=\) \(36.9^\circ\) and \(143.1^\circ\).
   2. Divide by 2 first; cosine is negative in the second and third quadrants, with related angle \(60^\circ\). \(\cos\theta=-\tfrac12\): \(120^\circ\) and \(240^\circ\).
   3. Tangent has period \(180^\circ\), so add multiples of \(180^\circ\) to the principal value. Principal value \(-63.4^\circ\), add \(180^\circ\) and \(360^\circ\): \(116.6^\circ\) and \(296.6^\circ\).
   4. Double the interval, because \(\theta\) from \(0^\circ\) to \(360^\circ\) means \(2\theta\) from \(0^\circ\) to \(720^\circ\). Let \(u=2\theta\) in the range \(0^\circ\le u\le720^\circ\). \(\sin u=0.5\) gives \(u=30^\circ,150^\circ,390^\circ,510^\circ\), so \(\theta=\) \(15^\circ,75^\circ,195^\circ,255^\circ\).

7. Answer to Question 7

   1. Two sides and the included angle are known, so use the cosine rule. \(AB^2=120^2+95^2-2\times120\times95\cos70^\circ=23425-22800\cos70^\circ=15627\), so \(AB=\) 125 m. The lake is about 125 m wide.
   2. Use area \(=\tfrac12ab\sin C\). Area \(=\tfrac12\times120\times95\times\sin70^\circ=\) 5 356 m\(^2\). The area is about 5 356 m\(^2\).
   3. Multiply the area by the price per square metre. \(5356\times14.50=77\,662\), so about £77 700. The land costs about £77 700.
   4. Use the sine rule with the side \(OB=95\) m opposite angle \(OAB\). \(\dfrac{\sin A}{95}=\dfrac{\sin70^\circ}{125.0}\), so \(\sin A=0.7141\) and \(A=\) 45.6\(^\circ\). Angle \(OAB\approx45.6^\circ\).

8. Answer to Question 8

   1. Draw north lines at \(P\) and at \(Q\); the bearing of \(P\) from \(Q\) is the back-bearing. At \(Q\), the back-bearing to \(P\) is \(040^\circ+180^\circ=220^\circ\). The ship\'s new direction is \(110^\circ\). The angle between them is \(220^\circ-110^\circ=110^\circ\).
   2. Use the cosine rule with \(PQ=20\), \(QR=30\) and the included angle \(110^\circ\). \(PR^2=20^2+30^2-2\times20\times30\cos110^\circ=1300+410.4=1710.4\), so \(PR=\) 41.4 km.
   3. Find angle \(QPR\) with the sine rule, then add it to the bearing of \(Q\) from \(P\). \(\dfrac{\sin P}{30}=\dfrac{\sin110^\circ}{41.36}\), so \(\sin P=0.6816\) and angle \(QPR=43.0^\circ\). Bearing \(=040^\circ+43^\circ=\) \(083^\circ\).
   4. Time \(=\dfrac{\text{distance}}{\text{speed}}\). Time \(=\dfrac{41.36}{18}=2.298\) h \(=2\) h \(18\) min (to the nearest minute).

9. Answer to Question 9

   1. Use angles on a straight line at \(B\) and then the angle sum of triangle \(ATB\). Angle \(TBA\) is \(180^\circ-41^\circ=139^\circ\) (angles on a straight line). Angle \(ATB=180^\circ-32^\circ-139^\circ=9^\circ\).
   2. Use the sine rule: side \(AT\) is opposite angle \(TBA=139^\circ\), and side \(AB=40\) m is opposite angle \(ATB=9^\circ\). \(\dfrac{AT}{\sin139^\circ}=\dfrac{40}{\sin9^\circ}\), so \(AT=\dfrac{40\sin139^\circ}{\sin9^\circ}=\) 167.8 m. So \(AT\approx167.8\) m.
   3. In the right-angled triangle formed by \(A\), the top and the base of the tower, the height is opposite the \(32^\circ\) angle. Height \(=AT\sin32^\circ=167.75\times0.5299=\) 89 m. The tower is about 89 m tall.
   4. In the right-angled triangle at \(B\), \(\tan41^\circ=\dfrac{\text{height}}{BH}\). Height \(=BH\tan41^\circ\) where \(BH\) is the distance, so \(BH=\dfrac{88.9}{\tan41^\circ}=\) 102 m. So \(B\) is about 102 m from the base of the tower.

10. Answer to Question 10

    1. Substitute \(t=3\) and \(t=9\) into the model. \(t=3\): \(4+2\sin90^\circ=\) 6 m. \(t=9\): \(4+2\sin270^\circ=\) 2 m. These are the high and low tide depths.
    2. Set \(h=5\) and solve for \(30t\) in the range \(0^\circ\) to \(360^\circ\). \(2\sin(30t)=1\), \(\sin(30t)=0.5\), \(30t=30^\circ\) or \(150^\circ\), so \(t=\) 1 and 5: 1 am and 5 am.
    3. Solve \(4+2\sin(30t)\ge4.5\); the sine is at least 0.25 between the two solutions in the first half-cycle. \(\sin(30t)\ge0.25\): \(30t\ge14.48^\circ\) and \(30t\le165.52^\circ\), so \(0.483\le t\le5.517\). That is from 12:29 am to 5:31 am.
    4. The period of \(\sin(kx^\circ)\) is \(\dfrac{360^\circ}k\). Period \(=\dfrac{360^\circ}{30^\circ}=12\) hours: the tide pattern repeats every 12 hours (a simplification of the real 12.4 hours).

11. Answer to Question 11

    1. In radians, the area of a sector is \(\tfrac12r^2\theta\). Area \(=\tfrac12r^2\theta=\tfrac12\times36\times\dfrac{2\pi}3=12\pi=\) 37.7 m\(^2\). The area is exactly \(12\pi\) m\(^2\).
    2. The arc length is \(r\theta\); add the two radii for the straight edges. Arc \(=r\theta=6\times\dfrac{2\pi}3=4\pi=12.57\) m. Perimeter \(=4\pi+6+6=\) 24.6 m. The perimeter is about 24.6 m.
    3. Multiply the perimeter by the cost per metre. Cost \(=24.57\times6.50=\) £159.68. The fencing costs about £159.68.
    4. Multiply the area by the cost per square metre. Cost \(=37.70\times4.20=\) £158.34. The turf costs about £158.34.

12. Answer to Question 12

    1. Split the isosceles triangle \(OAB\) into two right-angled triangles, each with angle \(0.6\) rad at \(O\); the calculator must be in radians. \(AB=2r\sin\dfrac\theta2=20\sin0.6=\) 11.29 cm. The chord is about 11.29 cm long.
    2. Use area \(=\tfrac12r^2\sin\theta\) with the angle between the two radii. Triangle \(=\tfrac12r^2\sin\theta=50\sin1.2=46.60\) cm\(^2\). The triangle has area about 46.60 cm\(^2\).
    3. The segment is the sector minus the triangle. Sector \(=\tfrac12\times100\times1.2=60\) cm\(^2\). Segment \(=60-46.60=\) 13.40 cm\(^2\). The segment has area about 13.40 cm\(^2\).
    4. Multiply the area by the price per cm\(^2\). Cost \(=13.40\times0.12=\) £1.61. The coloured glass costs about £1.61.

13. Answer to Question 13

    1. The largest angle is opposite the longest side; rearrange the cosine rule. The largest angle is opposite the longest side: \(\cos C=\dfrac{120^2+150^2-200^2}{2\times120\times150}=-0.0861\), so \(C=\) 94.9\(^\circ\). The negative cosine shows that the angle is obtuse.
    2. Use area \(=\tfrac12ab\sin C\) with the two sides enclosing the angle \(C\). Area \(=\tfrac12\times120\times150\sin94.94^\circ=\) 8 967 m\(^2\). The field has an area of about 8 967 m\(^2\).
    3. Heron's formula is \(\sqrt{s(s-a)(s-b)(s-c)}\) where \(s\) is half the perimeter. \(s=\dfrac{120+150+200}2=235\). Area \(=\sqrt{235\times115\times85\times35}=\sqrt{80\,393\,125}=8\,967\) m\(^2\) ✓. The two methods agree.
    4. Multiply the area by the price per square metre. Cost \(=8967\times0.45=£4\,035\), i.e. £4 040. The seed costs about £4 040.

14. Answer to Question 14

    1. Treat \(\sin x\) as a single variable and factorise the quadratic. Let \(s=\sin x\): \(2s^2+s-1=(2s-1)(s+1)\).
    2. Set each factor equal to zero and solve for \(x\) in the given range. \(\sin x=\tfrac12\) gives \(x=30^\circ\) or \(150^\circ\). \(\sin x=-1\) gives \(x=270^\circ\). Solutions: \(30^\circ,150^\circ,270^\circ\).
    3. \(\sin x\) can only take values from \(-1\) to 1, so find the greatest and least values of the quadratic on this interval. \(y=2s^2+s-1\) with \(-1\le s\le1\). At \(s=1\): \(y=2\) (maximum). The quadratic has its minimum at \(s=-\tfrac14\): \(y=2\times\tfrac1{16}-\tfrac14-1=-\tfrac98\) (minimum). Check the ends: at \(s=-1\), \(y=0\). So maximum 2, minimum \(-\tfrac98\).

15. Answer to Question 15

    1. In triangle \(ABC\) two sides and the included angle are known, so use the cosine rule. \(AC^2=8^2+6^2-2\times8\times6\cos75^\circ=100-96\cos75^\circ=75.15\), so \(AC=\) 8.67 m. The diagonal \(AC\) is about 8.67 m.
    2. In triangle \(ACD\) all three sides are now known, so rearrange the cosine rule. \(\cos D=\dfrac{9^2+5^2-75.15}{2\times9\times5}=\dfrac{30.85}{90}=0.3427\), so \(D=\) 70.0\(^\circ\). Angle \(ADC\approx70.0^\circ\).
    3. Find the area of each triangle with \(\tfrac12ab\sin C\) and add. Triangle \(ABC\): \(\tfrac12\times8\times6\sin75^\circ=23.18\). Triangle \(ACD\): \(\tfrac12\times9\times5\sin69.96^\circ=21.14\). Total \(=\) 44.3 m\(^2\). The plot has a total area of about 44.3 m\(^2\).
    4. Multiply the length of \(AC\) by the cost per metre. \(8.67\times11=\) £95.37. The fence costs about £95.37.

16. Answer to Question 16

    1. An equation with both \(\sin\) and \(\cos^2\) can be written entirely in \(\sin\theta\). Replace \(\cos^2\theta=1-\sin^2\theta\): \(2-2\sin^2\theta+5\sin\theta-4=0\), so \(2\sin^2\theta-5\sin\theta+2=0\), \((2\sin\theta-1)(\sin\theta-2)=0\). \(\sin\theta=2\) is impossible. \(\sin\theta=\tfrac12\) gives \(\theta=30^\circ,150^\circ\).
    2. Write \(\tan\theta=\dfrac{\sin\theta}{\cos\theta}\) and multiply by \(\cos\theta\). \(\dfrac{\sin\theta}{\cos\theta}=2\sin\theta\). Do not divide by \(\sin\theta\) (it would lose solutions). \(\sin\theta=2\sin\theta\cos\theta\), so \(\sin\theta(1-2\cos\theta)=0\). \(\sin\theta=0\): \(0^\circ,180^\circ,360^\circ\). \(\cos\theta=\tfrac12\): \(60^\circ,300^\circ\). Solutions: \(0^\circ,60^\circ,180^\circ,300^\circ,360^\circ\).
    3. Shift the interval to match the substitution. Let \(u=\theta+40^\circ\) with \(40^\circ\le u\le400^\circ\). \(\cos u=-\tfrac12\) gives \(u=120^\circ,240^\circ\). So \(\theta=\) \(80^\circ,200^\circ\).

17. Answer to Question 17

    1. The perimeter of a sector is two radii plus the arc. Perimeter \(=2r+r\theta=20\), so \(r\theta=20-2r\) and \(\theta=\dfrac{20-2r}r\).
    2. Substitute into \(\theta=\dfrac{20-2r}r\) and \(A=\tfrac12r^2\theta\). \(r=4\): \(\theta=\dfrac{12}4=3\), area \(=\tfrac12\times16\times3=24\) ✓. \(r=8\): \(\theta=\dfrac48=0.5\), area \(=\tfrac12\times64\times0.5=16\) ✓.
    3. Use \(A=\tfrac12r^2\theta\) and substitute \(r\theta=20-2r\). \(A=\tfrac12r^2\theta=\tfrac12r(r\theta)=\tfrac12r(20-2r)=r(10-r)\). This is the required result.
    4. \(A=10r-r^2=25-(r-5)^2\). Greatest area 25 cm\(^2\) at \(r=5\), where \(\theta=\dfrac{10}5=\) 2 rad. Check: perimeter \(=2\times5+5\times2=20\) ✓.

18. Answer to Question 18

    1. \(\sin2x\) varies between \(-1\) and 1, so \(3\sin2x\) varies between \(-3\) and 3. Maximum \(3+1=\) 4, minimum \(-3+1=\) \(-2\).
    2. The period of \(\sin kx\) is \(\dfrac{360^\circ}k\). Period \(=\dfrac{360^\circ}2=\) \(180^\circ\).
    3. Rearrange to make \(\sin2x\) the subject, then solve for \(2x\) in \(0^\circ\le2x\le720^\circ\). \(3\sin2x=1.5\), \(\sin2x=0.5\). \(2x=30^\circ,150^\circ,390^\circ,510^\circ\), so \(x=\) \(15^\circ,75^\circ,195^\circ,255^\circ\) (the points where the red line crosses the curve).
    4. A stretch with scale factor \(\tfrac12\) parallel to the \(x\)-axis, a stretch with scale factor 3 parallel to the \(y\)-axis and a translation by \(\begin{pmatrix}0\\1\end{pmatrix}\).

19. Answer to Question 19

    1. Use the difference of two squares \((1-\sin\theta)(1+\sin\theta)=1-\sin^2\theta=\cos^2\theta\). Multiply top and bottom of the left side by \((1+\sin\theta)\): \(\dfrac{\cos\theta(1+\sin\theta)}{1-\sin^2\theta}=\dfrac{\cos\theta(1+\sin\theta)}{\cos^2\theta}=\dfrac{1+\sin\theta}{\cos\theta}\) = RHS. This proves the identity.
    2. Add the fractions over the common denominator \((1+\sin\theta)(1-\sin\theta)\). LHS \(=\dfrac{(1-\sin\theta)+(1+\sin\theta)}{(1+\sin\theta)(1-\sin\theta)}=\dfrac2{1-\sin^2\theta}=\dfrac2{\cos^2\theta}\) = RHS. This proves the identity.
    3. Use the identity from part (b) to replace the left-hand side. \(\dfrac2{\cos^2\theta}=8\), so \(\cos^2\theta=\dfrac14\) and \(\cos\theta=\pm\dfrac12\). \(\cos\theta=\tfrac12\): \(60^\circ,300^\circ\). \(\cos\theta=-\tfrac12\): \(120^\circ,240^\circ\). Solutions: \(60^\circ,120^\circ,240^\circ,300^\circ\).

20. Answer to Question 20

    1. Use the sine rule to find \(\sin B\); since \(\sin B=\sin(180^\circ-B)\), check whether the obtuse angle also fits. \(\sin B=\dfrac{10\sin40^\circ}7=0.918\). So \(B=66.7^\circ\) or \(B=180^\circ-66.7^\circ=113.3^\circ\). Both give a valid angle sum with \(A=40^\circ\) (\(40^\circ+113.3^\circ<180^\circ\)), so two triangles exist.
    2. Find \(C\) from the angle sum, then use the sine rule for \(c\). Triangle 1: \(C=180^\circ-40^\circ-66.7^\circ=73.3^\circ\), \(c=\dfrac{7\sin73.3^\circ}{\sin40^\circ}=\) 10.4 cm. Triangle 2: \(C=180^\circ-40^\circ-113.3^\circ=26.7^\circ\), \(c=\dfrac{7\sin26.7^\circ}{\sin40^\circ}=\) 4.89 cm.
    3. The height of the triangle from \(C\) is \(b\sin A=10\sin40^\circ=6.43\). Two triangles exist when \(b\sin A\lt a\lt b\): \(6.43\lt a<10\). If \(a=6.43\) there is one right-angled triangle, if \(a<6.43\) none, and if \(a\ge10\) only one.