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

Applications of trigonometry Problem set A Worked solutions

Skills

  1. The cuboid ABCDEFGH has AB = 8 cm, BC = 5 cm and AE = 6 cm.

    8 cm5 cm6 cmABCDEFGH

    1. (a) Find the length of AG.

      • \(AG=\sqrt{8^2+5^2+6^2}\)
      • \(AG=11.2\text{ cm}\)

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

      • \(AC=\sqrt{8^2+5^2}=9.43\text{ cm}\)
      • \(\tan\theta=\frac{CG}{AC}=\frac{6}{9.43}\)
      • \(\theta=32.5^\circ\)

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

      • \(AB\perp BG, \quad BG=\sqrt{5^2+6^2}=7.81\text{ cm}\)
      • \(\tan\theta=\frac{BG}{AB}=\frac{7.81}{8}\)
      • \(\theta=44.3^\circ\)

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

      • \(\text{Let O be the centre. }AO=OC=5.59\text{ cm}, \ AC=9.43\text{ cm}\)
      • \(\cos AOC=\frac{5.59^2+5.59^2-9.43^2}{2(5.59)(5.59)}\)
      • \(\angle AOC=115^\circ, \text{ acute angle }=180^\circ-115^\circ=64.9^\circ\)

  2. The cuboid ABCDEFGH has AB = 10 cm, BC = 6 cm and AE = 5 cm.

    10 cm6 cm5 cmABCDEFGH

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

      • \(AC=\sqrt{10^2+6^2}=11.7\text{ cm}\)
      • \(\tan ACE=\frac{AE}{AC}=\frac{5}{11.7}\)
      • \(\angle ACE=23.2^\circ\)

    2. (b) Find the length of HB.

      • \(HB=\sqrt{10^2+6^2+5^2}=12.7\text{ cm}\)

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

      • \(BD=AC=11.7\text{ cm}, \quad HD=5\text{ cm}\)
      • \(\tan HBD=\frac{HD}{BD}=\frac{5}{11.7}\)
      • \(\angle HBD=23.2^\circ\)

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

      • \(BG=\sqrt{6^2+5^2}=7.81, \ BE=\sqrt{10^2+5^2}=11.2, \ EG=\sqrt{10^2+6^2}=11.7\)
      • \(\cos GBE=\frac{7.81^2+11.2^2-11.7^2}{2(7.81)(11.2)}, \quad \angle GBE=73.4^\circ\)
      • \(\text{Area}=\tfrac12(7.81)(11.2)\sin 73.4^\circ=41.8\text{ cm}^2\)

  3. In the cuboid ABCDEFGH the three face diagonals AC, AF and CF have lengths AC = 10 cm, AF = 8 cm and CF = 9 cm.

    10 cm8 cm9 cmABCDEFGH

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

      • \(AB^2+BC^2=10^2, \ AB^2+AE^2=8^2, \ BC^2+AE^2=9^2\)
      • \(2(AB^2+BC^2+AE^2)=10^2+8^2+9^2=245\)
      • \(AG^2=122.5, \quad AG=11.07\text{ cm}\)

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

      • \(AE^2=122.5-10^2=22.5, \quad AE=4.74\text{ cm}\)
      • \(\tan\theta=\frac{AE}{AC}=\frac{4.74}{10}\)
      • \(\theta=25.4^\circ\)

  4. EABCD is a right pyramid with a square base ABCD of side 8 cm. Each sloping edge is 11 cm long. M is the midpoint of AB and N is the midpoint of BC.

    11 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.
      • \(AC=\sqrt{8^2+8^2}=11.3\text{ cm}\)
      • Mark the height from E to the midpoint of AC, with a right angle.

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

      • \(\text{Half of }AC=5.66\text{ cm}\)
      • \(h=\sqrt{11^2-5.66^2}=9.43\text{ cm}\)

    3. (c) Find the angle AEC.

      • \(\sin\left(\tfrac12 AEC\right)=\frac{5.66}{11}\)
      • \(\angle AEC=61.9^\circ\)

    4. (d) Find the angle MEN.

      • \(EM=EN=\sqrt{11^2-4.00^2}=10.2\text{ cm}, \quad MN=\tfrac12 AC=5.66\text{ cm}\)
      • \(\cos MEN=\frac{2(10.2)^2-5.66^2}{2(10.2)^2}\)
      • \(\angle MEN=32.0^\circ\)

  5. MABCD is a right pyramid. The base ABCD is a rectangle with AB = 9 cm and BC = 6 cm. O is the centre of the base and OM = 7 cm is the height.

    9 cm6 cm7 cmABCMO

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

      • \(AC=\sqrt{9^2+6^2}=10.8\text{ cm}, \quad OC=5.41\text{ cm}\)
      • \(\tan ACM=\frac{OM}{OC}=\frac{7}{5.41}\)
      • \(\angle ACM=52.3^\circ\)

    2. (b) Find the angle ABM.

      • \(MB=\sqrt{7^2+5.41^2}=8.85\text{ cm}\)
      • Triangle ABM is isosceles with MA = MB. Split it in half: the foot of the perpendicular from M is the midpoint of AB.
      • \(\cos ABM=\frac{4.50}{8.85}, \quad \angle ABM=59.4^\circ\)

  6. Two solids.

    ABMO

    1. (a) MABC is a right pyramid. The base ABC is an equilateral triangle of side 6 cm. O is the centre of the base and OM = 5 cm. Find the angle between AM and the base.

      • \(\text{O is the centre of the triangle, so }OA=\frac{6}{\sqrt3}=3.46\text{ cm}\)
      • \(\tan MAO=\frac{5}{3.46}\)
      • \(\angle MAO=55.3^\circ\)

    2. (b) A cone has a base of diameter 10 cm and a vertical height of 7 cm. Find the angle between a sloping edge and the base.

      • \(\text{Radius}=5\text{ cm}\)
      • \(\tan\theta=\frac{7}{5}\)
      • \(\theta=54.5^\circ\)

In context

  1. A boat leaves a harbour H and sails 5.3 km on a bearing of 072° to a buoy K. It then sails 4.1 km on a bearing of 158° to a wreck W.

    5.3 km4.1 km?HKWNN

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

      • \(\angle HKW=180^\circ-(158^\circ-72^\circ)=94^\circ\)
      • \(HW^2=5.3^2+4.1^2-2(5.3)(4.1)\cos 94^\circ\)
      • \(HW=6.92\text{ km}\)

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

      • \(\cos H=\frac{5.3^2+6.92^2-4.1^2}{2(5.3)(6.92)}, \quad H=36.2^\circ\)
      • \(\text{Bearing of }W\text{ from }H=72^\circ+36.2^\circ=108^\circ\)
      • \(\text{Bearing back}=108^\circ+180^\circ=288°\)

  2. From a coastguard post C, a mast M is 3.6 km away on a bearing of 025° and a wreck W is 2.9 km away on a bearing of 112°.

    3.6 km2.9 kmCMWN

    1. (a) Find the distance from the mast to the wreck.

      • \(\text{Angle at C}=87.0^\circ\)
      • \(MW^2=3.6^2+2.9^2-2(3.6)(2.9)\cos 87.0^\circ\)
      • \(MW=4.50\text{ km}\)

    2. (b) Find the bearing of the mast from the wreck. Give the bearing to the nearest degree.

      • \(\cos W=\frac{2.9^2+4.50^2-3.6^2}{2(2.9)(4.50)}, \quad \angle CWM=53.0^\circ\)
      • \(\text{Bearing of C from W}=112^\circ+180^\circ=292^\circ\)
      • M is clockwise from the line WC as seen from W, so \(292^\circ+53.0^\circ=345^\circ\)

  3. A hut H has a pylon P 250 m away on a bearing of 125°. A well W is due east of the hut and due north of the pylon.

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

      • Draw the north line at H, then the pylon 125° clockwise from it.
      • Draw the east line from H and a vertical line from the pylon. They meet at the well, with a right angle.

    2. (b) Find the distance from the well to the pylon.

      • \(\text{Angle at H between the }east\text{ line and the pylon}=35.0^\circ\)
      • \(\text{Opposite}=\text{hypotenuse}\times\sin: \quad 250\sin 35.0^\circ\)
      • \(143\text{ m}\)

  4. A ranger walks 6.4 km on a bearing of 048°, then 3.5 km on a bearing of 126°, then 4.1 km on a bearing of 228°. Find how far she is from her start.

    6.4 km3.5 km4.1 km?SPQFN
    • \(\text{Legs 1 and 3 are on bearings }048^\circ\text{ and }228^\circ\text{, which are opposite. Net distance on }048^\circ=6.4-4.1=2.3\text{ km}\)
    • \(\text{Angle between the net displacement and leg 2: }180^\circ-(126^\circ-48^\circ)=102^\circ\)
    • \(d^2=2.3^2+3.5^2-2(2.3)(3.5)\cos 102^\circ\)
    • \(d=4.57\text{ km}\)

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

    1. (a) Priya's eyes are 1.5 m above the ground. She stands 18 m from a flagpole and sees the top at an angle of elevation of 34°. Find the height of the flagpole.

      • \(\text{Height above the eye}=18\tan 34^\circ=12.1\text{ m}\)
      • \(\text{Total}=1.5+12.1=13.6\text{ m}\)

    2. (b) From a point on level ground 55 m from the foot of a tower, the angle of elevation of the top of the tower is 28°. Find the height of the tower.

      • \(\tan 28^\circ=\frac{h}{55}\)
      • \(h=55\tan 28^\circ=29.2\text{ m}\)

    3. (c) The top of a crane 60 m high is seen at an angle of elevation of 25° from a point on level ground. Find the horizontal distance from the point to the crane.

      • \(\tan 25^\circ=\frac{60}{x}\)
      • \(x=\frac{60}{\tan 25^\circ}=129\text{ m}\)

    4. (d) From a balcony 14 m above the ground, the angle of depression of a car is 18°. Find the horizontal distance from the balcony to the car.

      • The angle of depression from the top equals the angle of elevation from the bottom (alternate angles).
      • \(\tan 18^\circ=\frac{14}{x}\)
      • \(x=\frac{14}{\tan 18^\circ}=43.1\text{ m}\)

  6. Three problems about heights and angles.

    1. (a) A clock tower stands on level ground. From a point on the ground the angle of elevation of the top is 36°. After walking 15 m directly away from the tower, the angle of elevation is 26°. Find the height of the tower.

      • \(\tan 36^\circ=\frac{h}{x}, \quad \tan 26^\circ=\frac{h}{x+15}\)
      • \(15=\frac{h}{\tan 26^\circ}-\frac{h}{\tan 36^\circ}\)
      • \(h=22.3\text{ m}\)

    2. (b) From the top of a watchtower, the angle of depression of a boat due east of the tower is 31.0°. A second boat is 25 m due south of the first. Its angle of depression from the top is 29.5°. Find the height of the watchtower.

      • \(\text{Let the height be }h. \text{ Nearer: }x=\frac{h}{\tan 31.0^\circ}. \text{ Farther: }y=\frac{h}{\tan 29.5^\circ}\)
      • \(y^2=x^2+25^2 \ \text{(right angle at the foot of the nearer one)}\)
      • \(h^2\left(\frac{1}{\tan^2 29.5^\circ}-\frac{1}{\tan^2 31.0^\circ}\right)=25^2\)
      • \(h=42.0\text{ m}\)

    3. (c) A tree stands at B on level ground. Priya stands at P, 26.0 m from B, and Omar stands at Q, 18.4 m from B. They are 15.2 m apart. Find the angle PBQ.

      • \(\cos \theta=\frac{26.0^2+18.4^2-15.2^2}{2(26.0)(18.4)}\)
      • \(\theta=35.0^\circ\)

Stretch

  1. A botanical garden has a glass greenhouse in the shape of a square-based pyramid. The base has sides of 28 m and the height is 17.5 m. Air conditioning is rated at one unit for every 600 m³ of air.

    28 m17.5 mABCEO

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

      • \(V=\tfrac13\times 28^2\times 17.5=4570\text{ m}^3\)
      • \(4570\div 600=7.62\)
      • Round up: 8 units. Assumptions: the greenhouse is a solid pyramid full of air, with nothing taking up space inside; units are whole numbers, so round up

    2. (b) On a hot day the glass gains energy at 150 watts for every m² of glass. Find the total power the air conditioning must remove. The base is not glass.

      • \(\text{Slant height}=\sqrt{17.5^2+14^2}=22.41\text{ m}\)
      • \(\text{Area of 4 faces}=4\times\tfrac12\times 28\times 22.41=1255\text{ m}^2\)
      • \(\text{Power}=1255\times 150\approx 188000\text{ W}=188\text{ kW}\)

    3. (c) The cleaning rules say scaffolding is needed if any glass has an angle of elevation of more than 48°. Is scaffolding needed?

      • \(\tan\theta=\frac{17.5}{14}\)
      • \(\theta=51.3^\circ\)
      • This is the steepest slope of a face, so it is the largest angle of elevation. 51.3° is more than 48°, so scaffolding is needed.

  2. A house is 9.0 m long and 6.4 m wide. Its roof is a triangular prism with an isosceles triangle as its end. Each side of the roof makes an angle of 38° with the horizontal. A loft conversion can use only the part of the roof space where the roof is at least 1.4 m above the floor.

    6.4 m38°

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

      • \(\tan 38^\circ=\frac{h}{3.2}\)
      • \(h=3.2\tan 38^\circ=2.50\text{ m}\)

    2. (b) Find the volume of the whole roof space.

      • \(V=\tfrac12\times 6.4\times 2.50\times 9=72.0\text{ m}^3\)

    3. (c) Find the percentage of the floor area that can be used.

      • \(\text{Height above the limit}=2.50-1.4=1.10\text{ m}\)
      • The usable triangle is similar to the whole end triangle. All its lengths are multiplied by
      • \(k=\frac{1.10}{2.50}=0.440\)
      • \(\text{Floor width scales by }k\text{, so the area is }44.0\%\)

    4. (d) Find the percentage of the roof space volume that can be used.

      • \(\text{Cross-section area scales by }k^2, \text{ the length is the same}\)
      • \(k^2=0.194\text{, so the volume is }19.4\%\)

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