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IB Mathematics: Applications and Interpretation SL · 12.10: Geometry & Trigonometry · 12.10.8

Applications of trigonometry Problem set B One question at a time

Try each question on paper first. Open the working a line at a time, then say whether you had it.

Skills

  1. The cuboid ABCDEFGH has AB = 7 cm, BC = 6 cm and AE = 4 cm.

    7 cm6 cm4 cmABCDEFGH

    1. (a) Find the length of AG.

      • \(AG=\sqrt{7^2+6^2+4^2}\)
      • \(AG=10.0\text{ cm}\)

    2. (b) Find the angle between AG and the base ABCD.

      • \(AC=\sqrt{7^2+6^2}=9.22\text{ cm}\)
      • \(\tan\theta=\frac{CG}{AC}=\frac{4}{9.22}\)
      • \(\theta=23.5^\circ\)

    3. (c) Find the angle between AG and the edge AB.

      • \(AB\perp BG, \quad BG=\sqrt{6^2+4^2}=7.21\text{ cm}\)
      • \(\tan\theta=\frac{BG}{AB}=\frac{7.21}{7}\)
      • \(\theta=45.9^\circ\)

    4. (d) AG and CE cross at the centre of the cuboid. Find the acute angle between AG and CE.

      • \(\text{Let O be the centre. }AO=OC=5.02\text{ cm}, \ AC=9.22\text{ cm}\)
      • \(\cos AOC=\frac{5.02^2+5.02^2-9.22^2}{2(5.02)(5.02)}\)
      • \(\angle AOC=133^\circ, \text{ acute angle }=180^\circ-133^\circ=46.9^\circ\)

  2. The cuboid ABCDEFGH has AB = 13 cm, BC = 5 cm and AE = 6 cm.

    13 cm5 cm6 cmABCDEFGH

    1. (a) Find AC, then find the angle ACE.

      • \(AC=\sqrt{13^2+5^2}=13.9\text{ cm}\)
      • \(\tan ACE=\frac{AE}{AC}=\frac{6}{13.9}\)
      • \(\angle ACE=23.3^\circ\)

    2. (b) Find the length of HB.

      • \(HB=\sqrt{13^2+5^2+6^2}=15.2\text{ cm}\)

    3. (c) Find the angle between HB and BD.

      • \(BD=AC=13.9\text{ cm}, \quad HD=6\text{ cm}\)
      • \(\tan HBD=\frac{HD}{BD}=\frac{6}{13.9}\)
      • \(\angle HBD=23.3^\circ\)

    4. (d) Find the area of the triangle BGE.

      • \(BG=\sqrt{5^2+6^2}=7.81, \ BE=\sqrt{13^2+6^2}=14.3, \ EG=\sqrt{13^2+5^2}=13.9\)
      • \(\cos GBE=\frac{7.81^2+14.3^2-13.9^2}{2(7.81)(14.3)}, \quad \angle GBE=71.2^\circ\)
      • \(\text{Area}=\tfrac12(7.81)(14.3)\sin 71.2^\circ=52.9\text{ cm}^2\)

  3. In the cuboid ABCDEFGH the three face diagonals AC, AF and CF have lengths AC = 12 cm, AF = 9 cm and CF = 10 cm.

    12 cm9 cm10 cmABCDEFGH

    1. (a) Find the length of the diagonal AG. Give your answer to 4 significant figures.

      • \(AB^2+BC^2=12^2, \ AB^2+AE^2=9^2, \ BC^2+AE^2=10^2\)
      • \(2(AB^2+BC^2+AE^2)=12^2+9^2+10^2=325\)
      • \(AG^2=162.5, \quad AG=12.75\text{ cm}\)

    2. (b) Find the angle between AG and the base ABCD.

      • \(AE^2=162.5-12^2=18.5, \quad AE=4.30\text{ cm}\)
      • \(\tan\theta=\frac{AE}{AC}=\frac{4.30}{12}\)
      • \(\theta=19.7^\circ\)

  4. EABCD is a right pyramid with a square base ABCD of side 6 cm. Each sloping edge is 9 cm long. M is the midpoint of AB and N is the midpoint of BC.

    9 cmABCEMN

    1. (a) Sketch the triangle ACE and label every length you can.

      • Draw triangle ACE on its own: AE = EC, with AC at the bottom.
      • \(AC=\sqrt{6^2+6^2}=8.49\text{ cm}\)
      • Mark the height from E to the midpoint of AC, with a right angle.

    2. (b) Find the height of the pyramid.

      • \(\text{Half of }AC=4.24\text{ cm}\)
      • \(h=\sqrt{9^2-4.24^2}=7.94\text{ cm}\)

    3. (c) Find the angle AEC.

      • \(\sin\left(\tfrac12 AEC\right)=\frac{4.24}{9}\)
      • \(\angle AEC=56.3^\circ\)

    4. (d) Find the angle MEN.

      • \(EM=EN=\sqrt{9^2-3.00^2}=8.49\text{ cm}, \quad MN=\tfrac12 AC=4.24\text{ cm}\)
      • \(\cos MEN=\frac{2(8.49)^2-4.24^2}{2(8.49)^2}\)
      • \(\angle MEN=29.0^\circ\)

  5. MABCD is a right pyramid. The base ABCD is a rectangle with AB = 12 cm and BC = 5 cm. O is the centre of the base and OM = 8 cm is the height.

    12 cm5 cm8 cmABCMO

    1. (a) Find the angle ACM, which is the angle between the sloping edge CM and the base.

      • \(AC=\sqrt{12^2+5^2}=13.0\text{ cm}, \quad OC=6.50\text{ cm}\)
      • \(\tan ACM=\frac{OM}{OC}=\frac{8}{6.50}\)
      • \(\angle ACM=50.9^\circ\)

    2. (b) Find the angle ABM.

      • \(MB=\sqrt{8^2+6.50^2}=10.3\text{ cm}\)
      • Triangle ABM is isosceles with MA = MB. Split it in half: the foot of the perpendicular from M is the midpoint of AB.
      • \(\cos ABM=\frac{6.00}{10.3}, \quad \angle ABM=54.4^\circ\)

  6. Two solids.

    ABMO

    1. (a) MABC is a right pyramid. The base ABC is an equilateral triangle of side 8 cm. O is the centre of the base and OM = 6 cm. Find the angle between AM and the base.

      • \(\text{O is the centre of the triangle, so }OA=\frac{8}{\sqrt3}=4.62\text{ cm}\)
      • \(\tan MAO=\frac{6}{4.62}\)
      • \(\angle MAO=52.4^\circ\)

    2. (b) A cone has a base of diameter 16 cm and a vertical height of 10 cm. Find the angle between a sloping edge and the base.

      • \(\text{Radius}=8\text{ cm}\)
      • \(\tan\theta=\frac{10}{8}\)
      • \(\theta=51.3^\circ\)

In context

  1. A hiker leaves a campsite C and walks 7.4 km on a bearing of 285° to a lake L. She then walks 5.2 km on a bearing of 210° to a hut H.

    7.4 km5.2 km?CLHNN

    1. (a) Find the distance from H back to C.

      • \(\angle CLH=180^\circ-(285^\circ-210^\circ)=105^\circ\)
      • \(CH^2=7.4^2+5.2^2-2(7.4)(5.2)\cos 105^\circ\)
      • \(CH=10.1\text{ km}\)

    2. (b) Find the bearing on which it must travel from H to return to C. Give the bearing to the nearest degree.

      • \(\cos C=\frac{7.4^2+10.1^2-5.2^2}{2(7.4)(10.1)}, \quad C=29.9^\circ\)
      • \(\text{Bearing of }H\text{ from }C=285^\circ-29.9^\circ=255^\circ\)
      • \(\text{Bearing back}=255^\circ-180^\circ=075°\)

  2. From a radar station R, a ship S is 5.2 km away on a bearing of 310° and a buoy B is 3.9 km away on a bearing of 047°.

    5.2 km3.9 kmRSBN

    1. (a) Find the distance from the ship to the buoy.

      • \(\text{Angle at R}=97.0^\circ\)
      • \(SB^2=5.2^2+3.9^2-2(5.2)(3.9)\cos 97.0^\circ\)
      • \(SB=6.87\text{ km}\)

    2. (b) Find the bearing of the ship from the buoy. Give the bearing to the nearest degree.

      • \(\cos B=\frac{3.9^2+6.87^2-5.2^2}{2(3.9)(6.87)}, \quad \angle RBS=48.7^\circ\)
      • \(\text{Bearing of R from B}=47^\circ+180^\circ=227^\circ\)
      • S is clockwise from the line BR as seen from B, so \(227^\circ+48.7^\circ=276^\circ\)

  3. A campsite C has a flagpole F 310 m away on a bearing of 308°. A stream S is due west of the campsite and due south of the flagpole.

    1. (a) Sketch the situation. Mark the north direction at the starting point, the bearing and the right angle.

      • Draw the north line at C, then the flagpole 308° clockwise from it.
      • Draw the west line from C and a vertical line from the flagpole. They meet at the stream, with a right angle.

    2. (b) Find the distance from the stream to the flagpole.

      • \(\text{Angle at C between the }west\text{ line and the flagpole}=38.0^\circ\)
      • \(\text{Opposite}=\text{hypotenuse}\times\sin: \quad 310\sin 38.0^\circ\)
      • \(191\text{ m}\)

  4. A drone flies 5.8 km on a bearing of 072°, then 2.9 km on a bearing of 158°, then 3.6 km on a bearing of 252°. Find how far it is from where it took off.

    5.8 km2.9 km3.6 km?SPQFN
    • \(\text{Legs 1 and 3 are on bearings }072^\circ\text{ and }252^\circ\text{, which are opposite. Net distance on }072^\circ=5.8-3.6=2.2\text{ km}\)
    • \(\text{Angle between the net displacement and leg 2: }180^\circ-(158^\circ-72^\circ)=94^\circ\)
    • \(d^2=2.2^2+2.9^2-2(2.2)(2.9)\cos 94^\circ\)
    • \(d=3.76\text{ km}\)

  5. Find the answer to each part. Draw a sketch with a right-angled triangle first.

    1. (a) A surveyor's eyes are 1.65 m above the ground. She stands 22 m from a tree and sees the top at an angle of elevation of 29°. Find the height of the tree.

      • \(\text{Height above the eye}=22\tan 29^\circ=12.2\text{ m}\)
      • \(\text{Total}=1.65+12.2=13.8\text{ m}\)

    2. (b) From a point on level ground 42 m from the foot of a cliff, the angle of elevation of the top of the cliff is 33°. Find the height of the cliff.

      • \(\tan 33^\circ=\frac{h}{42}\)
      • \(h=42\tan 33^\circ=27.3\text{ m}\)

    3. (c) A kite is 45 m above the ground. It is seen at an angle of elevation of 38° from a point on level ground. Find the horizontal distance from the point to the kite.

      • \(\tan 38^\circ=\frac{45}{x}\)
      • \(x=\frac{45}{\tan 38^\circ}=57.6\text{ m}\)

    4. (d) From the top of a cliff 85 m high, the angle of depression of a boat is 23°. Find the horizontal distance from the foot of the cliff to the boat.

      • The angle of depression from the top equals the angle of elevation from the bottom (alternate angles).
      • \(\tan 23^\circ=\frac{85}{x}\)
      • \(x=\frac{85}{\tan 23^\circ}=200\text{ m}\)

  6. Three problems about heights and angles.

    1. (a) A lamp-post stands on level ground. From a point on the ground the angle of elevation of the top is 41°. After walking 6 m directly away from the lamp-post, the angle of elevation is 30°. Find the height of the lamp-post.

      • \(\tan 41^\circ=\frac{h}{x}, \quad \tan 30^\circ=\frac{h}{x+6}\)
      • \(6=\frac{h}{\tan 30^\circ}-\frac{h}{\tan 41^\circ}\)
      • \(h=10.3\text{ m}\)

    2. (b) From the top of a ski-lift tower, the angle of depression of a skier due east of the tower is 27.0°. A second skier is 30 m due south of the first. Her angle of depression from the top is 25.0°. Find the height of the tower.

      • \(\text{Let the height be }h. \text{ Nearer: }x=\frac{h}{\tan 27.0^\circ}. \text{ Farther: }y=\frac{h}{\tan 25.0^\circ}\)
      • \(y^2=x^2+30^2 \ \text{(right angle at the foot of the nearer one)}\)
      • \(h^2\left(\frac{1}{\tan^2 25.0^\circ}-\frac{1}{\tan^2 27.0^\circ}\right)=30^2\)
      • \(h=34.7\text{ m}\)

    3. (c) A flagpole stands at B on level ground. Ana stands at P, 31.5 m from B, and Sam stands at Q, 22.8 m from B. They are 11.9 m apart. Find the angle PBQ.

      • \(\cos \theta=\frac{31.5^2+22.8^2-11.9^2}{2(31.5)(22.8)}\)
      • \(\theta=17.4^\circ\)

Stretch

  1. A shopping centre has an entrance in the shape of a glass square-based pyramid. The base has sides of 24 m and the height is 13 m. Heating is rated at one unit for every 900 m³ of air.

    24 m13 mABCEO

    1. (a) How many units are needed? State your assumptions.

      • \(V=\tfrac13\times 24^2\times 13=2500\text{ m}^3\)
      • \(2500\div 900=2.77\)
      • Round up: 3 units. Assumptions: the entrance is a solid pyramid full of air, with nothing taking up space inside; units are whole numbers, so round up

    2. (b) On a cold day the glass loses energy at 210 watts for every m² of glass. Find the total power the heating must supply. The base is not glass.

      • \(\text{Slant height}=\sqrt{13^2+12^2}=17.7\text{ m}\)
      • \(\text{Area of 4 faces}=4\times\tfrac12\times 24\times 17.7=849\text{ m}^2\)
      • \(\text{Power}=849\times 210=178000\text{ W}=178\text{ kW}\)

    3. (c) The cleaning rules say scaffolding is needed if any glass has an angle of elevation of more than 50°. Is scaffolding needed?

      • \(\tan\theta=\frac{13}{12}\)
      • \(\theta=47.3^\circ\)
      • This is the steepest slope of a face, so it is the largest angle of elevation. 47.3° is less than 50°, so scaffolding is not needed.

  2. A tent has a floor 3.8 m long and 2.6 m wide. Its roof is a triangular prism with an isosceles triangle as its end. Each side of the tent makes an angle of 62° with the ground. A person can stand only where the tent is at least 0.9 m high.

    2.6 m62°

    1. (a) Find the height of the ridge above the floor.

      • \(\tan 62^\circ=\frac{h}{1.3}\)
      • \(h=1.3\tan 62^\circ=2.44\text{ m}\)

    2. (b) Find the volume of air inside the tent.

      • \(V=\tfrac12\times 2.6\times 2.44\times 3.8=12.1\text{ m}^3\)

    3. (c) Find the percentage of the floor area where the tent is at least 0.9 m high.

      • \(\text{Height above the limit}=2.44-0.9=1.54\text{ m}\)
      • The usable triangle is similar to the whole end triangle. All its lengths are multiplied by
      • \(k=\frac{1.54}{2.44}=0.632\)
      • \(\text{Floor width scales by }k\text{, so the area is }63.2\%\)

    4. (d) Find the percentage of the volume of air that is at least 0.9 m above the floor.

      • \(\text{Cross-section area scales by }k^2, \text{ the length is the same}\)
      • \(k^2=0.399\text{, so the volume is }39.9\%\)

jmaths.xyz · AI SL · 12.10.8 Applications of trigonometry: 3D, bearings, elevation