Sign in to jmaths to print this page or save it as a PDF: jmaths-8ce638778fd6.herokuapp.com/maths/ai-sl/geometry-trigonometry/12-10-13/problems/

jmaths

IB Mathematics: Applications and Interpretation SL · 12.10: Geometry & Trigonometry · 12.10.13

Geometry and trigonometry: exam review Exam review Worked solutions

Section A: short questions

  1. A roof truss is a right-angled triangle ABC, with the right angle at B. The base AB is 8.4 m and the height BC is 2.9 m. [7]

    8.4 m2.9 mABC

    1. (a) Find the size of angle BAC. [2]

      • M1 for \(\tan A = \frac{2.9}{8.4}\)
      • A1 for 19.0°

    2. (b) A second roof truss has the same base AB and angle BAC = 25°. Find its height BC. [2]

      • M1 for \(\text{height} = 8.4 \tan 25^\circ\)
      • A1 for 3.92 m

    3. (c) Two straight roads on a map have equations \(y = 3x\) and \(y = -0.5x\). Find the acute angle between the roads. [3]

      • M1 for the angle of each line with the x-axis: \(\tan^{-1} 3 = 71.6°\) and \(\tan^{-1} 0.5 = 26.6°\)
      • M1 for adding the angles on either side of the x-axis: 71.6 + 26.6 = 98.1, then 180 − 98.1
      • A1 for 81.9°

  2. A triangular plot of land ABC has AB = 7.4 m, AC = 9.1 m and angle BAC = 64°. [7]

    7.4 m9.1 m64°ABC

    1. (a) Find the length of BC. [3]

      • M1 for \(BC^2 = 9.1^2 + 7.4^2 - 2(9.1)(7.4)\cos 64^\circ\)
      • A1 for BC² = 78.53
      • A1 for 8.86 m

    2. (b) Find the size of angle ABC. [2]

      • M1 for \(\frac{\sin B}{9.1} = \frac{\sin 64^\circ}{8.862}\)
      • A1 for 67.4°

    3. (c) Find the area of triangle ABC. [2]

      • M1 for \(\frac{1}{2}(9.1)(7.4)\sin 64^\circ\)
      • A1 for 30.3 m²

  3. A storage box ABCDEFGH is a cuboid with AB = 80 cm, BC = 50 cm and CG = 40 cm. [5]

    80 cm50 cm40 cmABCDEFGH

    1. (a) Find the length of AG. [2]

      • M1 for \(AG^2 = 80^2 + 50^2 + 40^2\)
      • A1 for 102 cm

    2. (b) Find the angle between AG and the base ABCD. [3]

      • M1 for \(AC = \sqrt{80^2 + 50^2} = 94.3\)
      • M1 for \(\tan \theta = \frac{40}{94.3}\) (angle GAC)
      • A1 for 23.0°

  4. A sector OAB of a circle has centre O, radius 9.5 cm and angle AOB = 72°. [5]

    72°9.5 cmOAB

    1. (a) Find the length of the arc AB. [2]

      • M1 for \(\frac{72}{360} \times 2\pi \times 9.5\)
      • A1 for 11.9 cm

    2. (b) Find the area of the segment between the chord AB and the arc AB. [3]

      • M1 for sector area \(\frac{72}{360} \times \pi \times 9.5^2 = 56.7\)
      • M1 for triangle area \(\frac{1}{2}(9.5)^2\sin 72^\circ = 42.9\)
      • A1 for 56.7 − 42.9 = 13.8 cm²

  5. Mia stands at P and measures the angle of elevation of the top T of a tower as 25°. She walks 45 m directly towards the tower to Q and measures the angle of elevation as 41°. Her eyes are 1.6 m above the ground. N is the point at the height of her eyes directly below T. [5]

    45 m25°41°PQNT

    1. (a) Find the size of angle PTQ. [1]

      • A1 for 41 − 25 = 16° (exterior angle of triangle PQT)

    2. (b) Use the sine rule in triangle PQT to find QT. [2]

      • M1 for \(\frac{QT}{\sin 25^\circ} = \frac{45}{\sin 16^\circ}\)
      • A1 for 69.0 m

    3. (c) Find the height of the tower above the ground. [2]

      • M1 for \(NT = 69.0 \sin 41^\circ = 45.3\)
      • A1 for 45.3 + 1.6 = 46.9 m

  6. A triangular garden ABC has AB = 9 m, AC = 12 m and area 40 m². Angle BAC is acute. The point D lies on BC with BD = 5 m. [7]

    1. (a) Find the size of angle BAC. [2]

      • M1 for \(40 = \frac{1}{2}(9)(12)\sin BAC\)
      • A1 for 47.8°

    2. (b) Find the length of BC. [2]

      • M1 for \(BC^2 = 9^2 + 12^2 - 2(9)(12)\cos 47.8^\circ\)
      • A1 for 8.94 m

    3. (c) Find the length of AD. [3]

      • M1 for \(\cos ABC = \frac{9^2 + 8.938^2 - 12^2}{2(9)(8.938)}\), giving angle ABC = 84.0°
      • M1 for \(AD^2 = 9^2 + 5^2 - 2(9)(5)\cos 84.0^\circ\)
      • A1 for 9.83 m

Section B: long questions

  1. A goat is tied to the corner P of a rectangular shed, 12 m by 8 m, standing on flat grass. The diagram shows the shed from above. The goat cannot go into the shed. [12]

    8 m12 mPM

    1. (a) The goat is tied at P by a rope of length 7 m. Find the area of grass the goat can reach. [3]

      • M1 for the angle outside the corner: \(360^\circ - 90^\circ = 270^\circ\), or \(\frac{3}{4}\) of a circle
      • M1 for \(\frac{3}{4} \times \pi \times 7^2\)
      • A1 for 115 m²

    2. (b) The rope is now 18 m long. After the rope has passed the two corners of the shed, find the radius of each of the two quarter circles the goat can reach beyond them. [2]

      • A1 for 18 − 12 = 6 m
      • A1 for 18 − 8 = 10 m

    3. (c) Find the total area of grass the goat can reach with the 18 m rope. [3]

      • M1 for \(\frac{3}{4}\pi(18)^2 = 763\)
      • M1 for the two quarter circles \(\frac{1}{4}\pi(6)^2 + \frac{1}{4}\pi(10)^2\)
      • A1 for 870 m²

    4. (d) The rope is moved to M, the midpoint of the 12 m wall, and shortened to 10 m. Find the area of grass the goat can now reach. [4]

      • M1 for a semicircle of radius 10: \(\frac{1}{2}\pi(10)^2 = 157\)
      • M1 for the rope that is left after 6 m along the wall: 10 − 6 = 4 m
      • M1 for two quarter circles of radius 4: \(2 \times \frac{1}{4}\pi(4)^2 = 25.1\)
      • A1 for 182 m²

  2. A rectangular field ABCD has AB = 25 m and AD = 15 m. A goat is tied at A by a rope of length 20 m. The field has a fence on all four sides, and the rope is long enough to reach the side DC. [12]

    25 m15 m20 mABCDEF

    1. (a) The rope just reaches the side DC at E, and AE = 20 m. Find the length DE and the size of angle EAB. [3]

      • M1 for \(DE = \sqrt{20^2 - 15^2}\)
      • A1 for DE = 13.2 m
      • A1 for \(\sin EAB = \frac{15}{20}\), angle EAB = 48.6°

    2. (b) The goat can reach the region ADE and the sector EAF, where F is on AB and AF = AE. Find the total area of the region the goat can reach. [4]

      • M1 for triangle ADE: \(\frac{1}{2}(15)(13.23)\)
      • A1 for 99.2 m²
      • M1 for sector EAF: \(\frac{48.59}{360} \times \pi \times 20^2\)
      • A1 for 99.2 + 170 = 269 m²

    3. (c) Find the percentage of the field that the goat can reach. [2]

      • M1 for \(\frac{269}{25 \times 15} \times 100\)
      • A1 for 71.7%

    4. (d) The rope length r m is changed, where 15 < r < 25. The goat can then reach half of the field. Write down an equation for r and solve it using technology. [3]

      • M1 for area as a function of r: \(\frac{1}{2}(15)\sqrt{r^2 - 15^2} + \frac{\sin^{-1}(15/r)}{360}\pi r^2\) (angle in degrees)
      • M1 for setting it equal to half the field, 188
      • A1 for r = 15.5

  3. A boat leaves port P and sails 14 km on a bearing of 052° to Q. It then sails 9 km on a bearing of 140° to R. [12]

    14 km9 km052°140°PQRNN

    1. (a) Find the size of angle PQR. [2]

      • M1 for the back-bearing of QP: 52 + 180 = 232
      • A1 for 232 − 140 = 92°

    2. (b) Find the distance PR. [3]

      • M1 for \(PR^2 = 14^2 + 9^2 - 2(14)(9)\cos 92^\circ\)
      • A1 for \(PR^2 = 285.8\)
      • A1 for 16.9 km

    3. (c) Find the size of angle QPR. [3]

      • M1 for \(\frac{\sin QPR}{9} = \frac{\sin 92^\circ}{16.91}\)
      • A1 for \(\sin QPR = 0.5320\)
      • A1 for 32.1°

    4. (d) Find the bearing of P from R. Give your answer to the nearest degree. [3]

      • M1 for the bearing of R from P: 52 + 32.14 = 84.14°
      • M1 for adding 180°
      • A1 for 264°

    5. (e) Write down the bearing of Q from R. [1]

      • A1 for 140 + 180 = 320°

  4. A glass pyramid VABCD stands in a museum courtyard. The base ABCD is a horizontal square of side 12 m and each sloping edge is 14 m. M is the centre of the base, and V is directly above M. [12]

    12 m14 mABCDMVN

    1. (a) Find the length of AC. [2]

      • M1 for \(AC^2 = 12^2 + 12^2\)
      • A1 for 17.0 m

    2. (b) Find the height VM of the pyramid. [2]

      • M1 for \(VM^2 = 14^2 - (8.485)^2\)
      • A1 for 11.1 m

    3. (c) Find the angle between the edge VA and the base ABCD. [2]

      • M1 for \(\cos VAM = \frac{8.485}{14}\) (or \(\tan VAM = \frac{11.14}{8.485}\))
      • A1 for 52.7°

    4. (d) N is the midpoint of AB. Find the angle between the face VAB and the base. [2]

      • M1 for \(\tan VNM = \frac{11.14}{6}\)
      • A1 for 61.7°

    5. (e) Find the volume of the pyramid. [2]

      • M1 for \(\frac{1}{3} \times 12^2 \times 11.14\)
      • A1 for 535 m³

    6. (f) Find the total area of the four triangular faces. [2]

      • M1 for the slant height \(VN = \sqrt{11.14^2 + 6^2} = 12.6\)
      • A1 for \(4 \times \frac{1}{2} \times 12 \times 12.65 = 304\) m²

  5. A glass is a cone with its point at the bottom. The radius of the top is 5 cm and the perpendicular height is 12 cm. The glass is held with its axis vertical and filled with juice to a depth of 7.2 cm. [11]

    1. (a) Find the volume of the glass when it is full, in cm³ and in litres. [2]

      • M1 for \(\frac{1}{3}\pi(5)^2(12)\)
      • A1 for 314 cm³ = 0.314 litres

    2. (b) Find the radius of the surface of the liquid. [2]

      • M1 for \(\frac{r}{5} = \frac{7.2}{12}\) (similar triangles)
      • A1 for 3 cm

    3. (c) Find the volume of the liquid, and write it as a percentage of the volume of the full glass. [3]

      • M1 for \(\frac{1}{3}\pi(3)^2(7.2)\)
      • A1 for 67.9 cm³
      • A1 for 21.6% (ft)

    4. (d) The liquid is poured into a cylinder with radius 4 cm. Find the depth of the liquid in the cylinder. [2]

      • M1 for \(\pi(4)^2 h = 67.9\)
      • A1 for 1.35 cm

    5. (e) Instead, the liquid is frozen into a solid hemisphere. Find the radius of the hemisphere. [2]

      • M1 for \(\frac{2}{3}\pi r^3 = 67.9\)
      • A1 for 3.19 cm

jmaths.xyz · AI SL · 12.10.13 Unit review