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IB Mathematics: Applications and Interpretation HL · 12.10: Geometry & Trigonometry · 12.10.4

Rules of trigonometry: mixed practice Problem set A Worked solutions

Skills

  1. The line \(L_1\) has equation \(3x+4y=24\). The line \(L_2\) has equation \(x-2y=8\).

    1. (a) Find the coordinates of the points where \(L_1\) meets the axes.

      • y = 0: \(3x=24\), so x = 8
      • x = 0: \(4y=24\), so y = 6

    2. (b) Find the acute angle between \(L_1\) and the x-axis.

      • Gradient of \(L_1\) = \(-\frac{3}{4}\)
      • \(\tan\theta=\frac{3}{4}\) (use the size of the gradient)
      • \(\theta\) = 36.9°

    3. (c) Show that \(L_1\) and \(L_2\) meet on the x-axis.

      • \(L_1\) at y = 0: x = 8, the point (8, 0)
      • \(L_2\) at y = 0: \(x=8\), so x = 8
      • Both lines cross the x-axis at the same point, so they meet on the x-axis

    4. (d) Find the acute angle between \(L_1\) and \(L_2\).

      • \(L_2\) has gradient \(\frac{1}{2}\) (positive): angle with the x-axis 26.6°
      • \(L_1\) has gradient negative: angle with the x-axis 36.9°
      • The lines slope opposite ways: 36.9° + 26.6° = 63.4°

  2. In the diagram, angle ABC is a right angle and D lies on BC. BD = 6 cm, angle BAD = 28° and angle DAC = 22°. Find the length x = DC.

    6 cmx28°, 22°ABDC
    • \(AB=\frac{6}{\tan 28^\circ}=11.28\) cm
    • \(BC=AB\tan 50^\circ=13.45\) cm (the angle at A in the big triangle is 28° + 22° = 50°)
    • \(x=BC-BD=13.45-6=7.45\) cm

  3. In the diagram, angle ABC is a right angle and D lies on BC. BD = 7 cm, DC = 3 cm and angle ACB = 24°. Find the size of angle BAD, marked θ.

    7 cm3 cmθ24°ABDC
    • BC = 7 + 3 = 10 cm
    • \(AB=BC\tan 24^\circ=4.452\) cm
    • \(\tan\theta=\frac{BD}{AB}=\frac{7}{4.452}\)
    • \(\theta=57.5^\circ\)

  4. In triangle ABC, AB = 6.1 cm, AC = 8.4 cm and angle BAC = 57°. Find the length of BC.

    6.1 cm8.4 cmx57°ABC
    • \(x^2=8.4^2+6.1^2-2\times8.4\times6.1\times\cos 57^\circ\)
    • \(x^2=51.96\)
    • \(x=7.21\) cm

  5. In triangle ABC, AB = 8.3 cm, BC = 9.6 cm and CA = 7.1 cm.

    8.3 cm9.6 cm7.1 cmθBCA

    1. (a) Find the size of angle ABC, marked θ.

      • \(\cos\theta=\frac{8.3^2+9.6^2-7.1^2}{2\times8.3\times9.6}=0.6943\)
      • \(\theta=46.0^\circ\)

    2. (b) Find the area of the triangle.

      • \(\text{Area}=\frac12\times8.3\times9.6\times\sin 46.03^\circ\)
      • = 28.7 cm²

  6. In triangle PQR, angle RPQ = 35°, QR = 7.5 cm and PR = 9 cm. There are two possible triangles. The diagram shows one.

    9 cm7.5 cm35°PRQ

    1. (a) Find the two possible sizes of angle PQR.

      • \(\frac{\sin Q}{9}=\frac{\sin 35^\circ}{7.5}\), so \(\sin Q=0.6883\)
      • First value: 43.5°
      • Second value: 180° − 43.5° = 137°. Check: 35° + 137° < 180°, so it is possible

    2. (b) Find the two possible sizes of angle PRQ.

      • Angles add to 180°: R = 180° − 35° − 43.5° = 102°
      • or R = 180° − 35° − 137° = 8.50°

In context

  1. Two coastguard stations, A and B, are 9.6 km apart on a straight, level coast. A ship S is at sea. From A the ship is seen at an angle of 52° to the coast line AB, and from B at an angle of 71° to BA.

    9.6 kmyd52°71°ABS

    1. (a) Find the distance y = AS from station A to the ship.

      • Third angle: 180° − 52° − 71° = 57.0°
      • \(\frac{y}{\sin 71^\circ}=\frac{9.6}{\sin 57.0^\circ}\)
      • y = 10.8 km

    2. (b) Find the shortest distance d from the ship to the coast.

      • The dashed line is perpendicular to AB: \(d=y\sin 52^\circ\)
      • d = 10.82 × sin 52° = 8.53 km

  2. A triangular garden ABC has AB = 12.4 m, AC = 9.7 m and angle BAC = 64°. A fence runs along BC.

    12.4 m9.7 mx64°θABC

    1. (a) Find the length x of the fence BC.

      • \(x^2=9.7^2+12.4^2-2\times9.7\times12.4\times\cos 64^\circ=142.4\)
      • x = 11.9 m

    2. (b) Find the size of angle ABC, marked θ.

      • \(\frac{\sin\theta}{9.7}=\frac{\sin 64^\circ}{11.93}\)
      • \(\sin\theta=0.7306\)
      • θ = 46.9° (θ is opposite the shorter side 9.7, so it is acute)

  3. A triangular sail ABC has area 14.5 m². The edge AB is 6.2 m long and the angle at A is 62°. The edge AC has length x and the edge BC has length y.

    6.2 mxy62°ABC

    1. (a) Find x.

      • \(14.5=\frac12\times6.2\times x\times\sin 62^\circ\)
      • \(x=\frac{2\times14.5}{6.2\sin 62^\circ}=5.30\) m

    2. (b) Find y.

      • \(y^2=6.2^2+5.298^2-2\times6.2\times5.298\times\cos 62^\circ=35.66\)
      • y = 5.97 m

Stretch

  1. In triangle ABC, D lies on BC. AB = 11 cm, AC = 14 cm, AD = 8 cm and angle ADB = 65°. Find the length of BC.

    11 cm14 cm8 cmx65°BCDA
    • Triangle ABD: \(11^2=8^2+BD^2-2\times8\times BD\times\cos 65^\circ\), so \(BD^2-6.762BD-57.00=0\) and BD = 11.65 cm
    • Angle ADC = 180° − 65° = 115°
    • Triangle ADC: \(14^2=8^2+DC^2-2\times8\times DC\times\cos 115^\circ\), so \(DC^2+6.762DC-132.0=0\) and DC = 8.595 cm
    • BC = 11.65 + 8.595 = 20.2 cm

  2. The area of triangle PQR is 28 cm². PQ = 7.4 cm, PR = 9.6 cm and angle QPR, marked θ, is acute.

    7.4 cm9.6 cmxθPQR

    1. (a) Find the size of θ.

      • \(28=\frac12\times7.4\times9.6\times\sin\theta\)
      • \(\sin\theta=0.7883\)
      • \(\theta=52.0^\circ\) (acute)

    2. (b) Find the length of QR.

      • \(QR^2=7.4^2+9.6^2-2\times7.4\times9.6\times\cos 52.03^\circ=59.50\)
      • QR = 7.71 cm

  3. In the diagram, D lies on BC. AB = 9 cm, angle ABD = 52°, angle ADB = 71° and angle ACD = 30°. Find the value of x = AC.

    9 cmx52°30°71°BDCA
    • Triangle ABD: \(\frac{AD}{\sin 52^\circ}=\frac{9}{\sin 71^\circ}\), so AD = 7.501 cm
    • Angle ADC = 180° − 71° = 109°
    • Triangle ADC: \(\frac{x}{\sin 109^\circ}=\frac{7.501}{\sin 30^\circ}\)
    • \(x=14.2\) cm

  4. A flagpole BT stands on the edge of a vertical cliff. The foot of the cliff is O, with OB = h metres and BT = d metres. From a point P on level ground, the angle of elevation of the bottom of the flagpole, B, is 32° and the angle of elevation of the top, T, is 36°.

    hd32°, 36°PBT

    1. (a) Show that \(h=\dfrac{d\tan 32^\circ}{\tan 36^\circ-\tan 32^\circ}\).

      • \(\tan 32^\circ=\frac{h}{PO}\), so \(PO=\frac{h}{\tan 32^\circ}\)
      • \(\tan 36^\circ=\frac{h+d}{PO}\), so \(PO=\frac{h+d}{\tan 36^\circ}\)
      • \(\frac{h}{\tan 32^\circ}=\frac{h+d}{\tan 36^\circ}\), so \(h\tan 36^\circ=(h+d)\tan 32^\circ\)
      • \(h(\tan 36^\circ-\tan 32^\circ)=d\tan 32^\circ\), so \(h=\frac{d\tan 32^\circ}{\tan 36^\circ-\tan 32^\circ}\)

    2. (b) The flagpole is 6 m long. Find the height of the cliff.

      • \(h=\frac{6\tan 32^\circ}{\tan 36^\circ-\tan 32^\circ}=36.9\) m

  5. A vertical floodlight mast OT of height h metres stands between two points P and Q on level ground. P, O and Q are in a straight line and PQ = 75 m. The angle of elevation of the top of the mast T from P is 27° and from Q is 48°.

    75 mh27°48°PQT

    1. (a) Write down PO and OQ in terms of h.

      • In triangle POT: \(PO=\frac{h}{\tan 27^\circ}\)
      • In triangle QOT: \(OQ=\frac{h}{\tan 48^\circ}\)

    2. (b) Hence find the value of h.

      • PO + OQ = 75, so \(\frac{h}{\tan 27^\circ}+\frac{h}{\tan 48^\circ}=75\)
      • Solve for h: h = 26.2 m

jmaths.xyz · AI HL · 12.10.4 Mixed practice: rules of trigonometry