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IB Mathematics: Applications and Interpretation HL · 12.10: Geometry & Trigonometry · 12.10.16

Geometry and trigonometry: exam review Exam review: second paper Worked solutions

Section A: short questions

  1. A road climbs a hillside from A to C. ABC is a right-angled triangle, with the right angle at B. The horizontal distance AB is 5.6 km and the height BC is 1.9 km. [7]

    5.6 km1.9 kmABC

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

      • M1 for \(\tan A = \frac{1.9}{5.6}\)
      • A1 for 18.7°

    2. (b) A second road has the same horizontal distance AB and rises at an angle BAC = 12°. Find its height BC. [2]

      • M1 for \(\text{height} = 5.6 \tan 12^\circ\)
      • A1 for 1.19 km

    3. (c) Two straight roads on a map have equations \(y = 2x\) and \(y = -1.5x\). Find the acute angle between the roads. [3]

      • M1 for the angle of each line with the x-axis: \(\tan^{-1} 2 = 63.4°\) and \(\tan^{-1} 1.5 = 56.3°\)
      • M1 for adding the angles on either side of the x-axis: 63.4 + 56.3 = 120, then 180 − 120
      • A1 for 60.3°

  2. A triangular sail PQR has PQ = 5.3 m, PR = 6.8 m and angle QPR = 58°. [7]

    5.3 m6.8 m58°PQR

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

      • M1 for \(QR^2 = 6.8^2 + 5.3^2 - 2(6.8)(5.3)\cos 58^\circ\)
      • A1 for QR² = 36.13
      • A1 for 6.01 m

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

      • M1 for \(\frac{\sin Q}{6.8} = \frac{\sin 58^\circ}{6.011}\)
      • A1 for 73.6°

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

      • M1 for \(\frac{1}{2}(6.8)(5.3)\sin 58^\circ\)
      • A1 for 15.3 m²

  3. A shipping crate ABCDEFGH is a cuboid with AB = 1.2 m, BC = 0.9 m and CG = 0.7 m. [5]

    1.2 m0.9 m0.7 mABCDEFGH

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

      • M1 for \(AG^2 = 1.2^2 + 0.9^2 + 0.7^2\)
      • A1 for 1.66 m

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

      • M1 for \(AC = \sqrt{1.2^2 + 0.9^2} = 1.50\)
      • M1 for \(\tan \theta = \frac{0.7}{1.50}\) (angle GAC)
      • A1 for 25.0°

  4. A slice of cake is a sector of a circle with radius 14 cm and angle 0.9 radians at the centre. [7]

    0.9 rad14 cmOAB

    1. (a) Write 0.9 radians in degrees, and write 135° in radians as a multiple of \(\pi\). [3]

      • M1 for \(0.9 \times \frac{180}{\pi}\) or \(135 \times \frac{\pi}{180}\)
      • A1 for 51.6°
      • A1 for \(\frac{3\pi}{4}\)

    2. (b) Find the arc length of the sector. [2]

      • M1 for \(s = r\theta = 14 \times 0.9\)
      • A1 for 12.6 cm

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

      • M1 for \(A = \frac{1}{2}r^2\theta = \frac{1}{2}(14)^2(0.9)\)
      • A1 for 88.2 cm²

  5. A garden bed is a sector OAB of a circle with centre O, radius 6.8 m and angle AOB = 110°. [5]

    110°6.8 mOAB

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

      • M1 for \(\frac{110}{360} \times 2\pi \times 6.8\)
      • A1 for 13.1 m

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

      • M1 for sector area \(\frac{110}{360} \times \pi \times 6.8^2 = 44.4\)
      • M1 for triangle area \(\frac{1}{2}(6.8)^2\sin 110^\circ = 21.7\)
      • A1 for 44.4 − 21.7 = 22.7 m²

  6. In triangle PQR, angle P = 35°, QR = 8.4 cm and PR = 12.5 cm. [6]

    12.5 cm8.4 cm35°PQR

    1. (a) Calculate 12.5 sin 35° and hence explain why there are two possible triangles. [2]

      • A1 for \(12.5 \sin 35^\circ = 7.17\)
      • R1 for 7.17 < 8.4 < 12.5, so side QR can meet the base in two places

    2. (b) Find the two possible values of angle PQR. [4]

      • M1 for \(\frac{\sin Q}{12.5} = \frac{\sin 35^\circ}{8.4}\)
      • A1 for 58.6°
      • M1 for \(180^\circ - 58.6^\circ\)
      • A1 for 121°

  7. A surveyor stands at P and measures the angle of elevation of the top T of a wind turbine as 18°. He walks 60 m directly towards the turbine to Q and measures the angle of elevation as 29°. His eyes are 1.7 m above the ground. N is the point at the height of his eyes directly below T. [5]

    60 m18°29°PQNT

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

      • A1 for 29 − 18 = 11° (exterior angle of triangle PQT)

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

      • M1 for \(\frac{QT}{\sin 18^\circ} = \frac{60}{\sin 11^\circ}\)
      • A1 for 97.2 m

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

      • M1 for \(NT = 97.2 \sin 29^\circ = 47.1\)
      • A1 for 47.1 + 1.7 = 48.8 m

  8. A triangular sail PQR has PQ = 11 m, PR = 8 m and area 35 m². Angle QPR is acute. The point S lies on QR with QS = 6 m. [7]

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

      • M1 for \(35 = \frac{1}{2}(11)(8)\sin QPR\)
      • A1 for 52.7°

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

      • M1 for \(QR^2 = 11^2 + 8^2 - 2(11)(8)\cos 52.7^\circ\)
      • A1 for 8.85 m

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

      • M1 for \(\cos PQR = \frac{11^2 + 8.851^2 - 8^2}{2(11)(8.851)}\), giving angle PQR = 46.0°
      • M1 for \(PS^2 = 11^2 + 6^2 - 2(11)(6)\cos 46.0^\circ\)
      • A1 for 8.08 m

  9. The angle \(\theta\) is obtuse and \(\sin\theta = \frac{8}{17}\). [4]

    1. (a) Find the exact value of \(\cos\theta\). [2]

      • M1 for \(\cos^2\theta = 1 - \left(\frac{8}{17}\right)^2 = \frac{225}{289}\)
      • A1 for \(-\frac{15}{17}\) (negative because \(\theta\) is obtuse)

    2. (b) Find the exact value of \(\tan\theta\). [1]

      • A1 for \(\tan\theta = \frac{\sin\theta}{\cos\theta} = -\frac{8}{15}\)

    3. (c) Find the size of \(\theta\), in degrees, to the nearest degree. [1]

      • A1 for \(180^\circ - 28.1^\circ = 152^\circ\)

Section B: long questions

  1. A goat is tied to the corner P of a rectangular barn, 14 m by 9 m, standing on flat grass. The diagram shows the barn from above. The goat cannot go into the barn. [12]

    9 m14 mPM

    1. (a) The goat is tied at P by a rope of length 8 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 \(\frac{3}{4} \times \pi \times 8^2\)
      • A1 for 151 m²

    2. (b) The rope is now 20 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 20 − 14 = 6 m
      • A1 for 20 − 9 = 11 m

    3. (c) Find the total area of grass the goat can reach with the 20 m rope. [3]

      • M1 for \(\frac{3}{4}\pi(20)^2 = 942\)
      • M1 for the two quarter circles \(\frac{1}{4}\pi(6)^2 + \frac{1}{4}\pi(11)^2\)
      • A1 for 1070 m²

    4. (d) The rope is moved to M, the midpoint of the 14 m wall, and shortened to 11 m. Find the area of grass the goat can now reach. [4]

      • M1 for a semicircle of radius 11: \(\frac{1}{2}\pi(11)^2 = 190\)
      • M1 for the rope that is left after 7 m along the wall: 11 − 7 = 4 m
      • M1 for two quarter circles of radius 4: \(2 \times \frac{1}{4}\pi(4)^2 = 25.1\)
      • A1 for 215 m²

  2. A rectangular paddock ABCD has AB = 22 m and AD = 12 m. A goat is tied at A by a rope of length 17 m. The paddock has a fence on all four sides, and the rope is long enough to reach the side DC. [12]

    22 m12 m17 mABCDEF

    1. (a) The rope just reaches the side DC at E, and AE = 17 m. Find the length DE and the size of angle EAB. [3]

      • M1 for \(DE = \sqrt{17^2 - 12^2}\)
      • A1 for DE = 12.0 m
      • A1 for \(\sin EAB = \frac{12}{17}\), angle EAB = 44.9°

    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: \(\frac{1}{2}(12)(12.04)\)
      • A1 for 72.2 m²
      • M1 for sector EAF: \(\frac{44.90}{360} \times \pi \times 17^2\)
      • A1 for 72.2 + 113 = 185 m²

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

      • M1 for \(\frac{185}{22 \times 12} \times 100\)
      • A1 for 70.3%

    4. (d) The rope length r m is changed, where 12 < r < 22. 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: \(\frac{1}{2}(12)\sqrt{r^2 - 12^2} + \frac{\sin^{-1}(12/r)}{360}\pi r^2\) (angle in degrees)
      • M1 for setting it equal to half the field, 132
      • A1 for r = 13.2

  3. A hiker leaves camp P and walks 12.5 km on a bearing of 112° to Q. The hiker then walks 9.6 km on a bearing of 215° to R. [12]

    12.5 km9.6 km112°215°PQRNN

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

      • M1 for the back-bearing of QP: 112 + 180 = 292
      • A1 for 292 − 215 = 77°

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

      • M1 for \(PR^2 = 12.5^2 + 9.6^2 - 2(12.5)(9.6)\cos 77^\circ\)
      • A1 for \(PR^2 = 194.4\)
      • A1 for 13.9 km

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

      • M1 for \(\frac{\sin QPR}{9.6} = \frac{\sin 77^\circ}{13.94}\)
      • A1 for \(\sin QPR = 0.6708\)
      • A1 for 42.1°

    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: 112 + 42.13 = 154.1°
      • M1 for adding 180°
      • A1 for 334°

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

      • A1 for 215 + 180 = 35°

  4. A tent VABCD is in the shape of a pyramid. The base ABCD is a horizontal square of side 10 m and each sloping edge is 13 m. M is the centre of the base, and V is directly above M. [12]

    10 m13 mABCDMVN

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

      • M1 for \(AC^2 = 10^2 + 10^2\)
      • A1 for 14.1 m

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

      • M1 for \(VM^2 = 13^2 - (7.071)^2\)
      • A1 for 10.9 m

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

      • M1 for \(\cos VAM = \frac{7.071}{13}\) (or \(\tan VAM = \frac{10.91}{7.071}\))
      • A1 for 57.0°

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

      • M1 for \(\tan VNM = \frac{10.91}{5}\)
      • A1 for 65.4°

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

      • M1 for \(\frac{1}{3} \times 10^2 \times 10.91\)
      • A1 for 364 m³

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

      • M1 for the slant height \(VN = \sqrt{10.91^2 + 5^2} = 12.0\)
      • A1 for \(4 \times \frac{1}{2} \times 10 \times 12.00 = 240\) m²

  5. The height of a pod on a wheel, \(h\) metres, \(t\) minutes after the wheel starts is modelled by \(h = 5.5 + 3\sin(0.45t)\), for \(0 \le t \le 12\), where the angle is in radians. [8]

    1. (a) Write down the greatest height and the least height. [2]

      • A1 for 8.5 m
      • A1 for 2.5 m

    2. (b) Find the period of the model. [1]

      • A1 for \(\frac{2\pi}{0.45} = 14.0\) minutes

    3. (c) Find the times, in the first 12 minutes, when the height is 6.5 m. [3]

      • M1 for \(5.5 + 3\sin(0.45t) = 6.5\) (or the graphs of \(h\) and \(h = 6.5\), radian mode)
      • A1 for t = 0.755
      • A1 for t = 6.23

    4. (d) Find the length of time, in the first 12 minutes, for which the height is at least 6.5 m. [2]

      • M1 for \(6.23 - 0.755\) (the height is above 6.5 m between the two times)
      • A1 for 5.47 minutes (accept 5.48)

  6. A paper cup is a cone with its point at the bottom. The radius of the top is 6 cm and the perpendicular height is 15 cm. The cup is held with its axis vertical and filled with water to a depth of 10 cm. [11]

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

      • M1 for \(\frac{1}{3}\pi(6)^2(15)\)
      • A1 for 565 cm³ = 0.565 litres

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

      • M1 for \(\frac{r}{6} = \frac{10}{15}\) (similar triangles)
      • A1 for 4 cm

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

      • M1 for \(\frac{1}{3}\pi(4)^2(10)\)
      • A1 for 168 cm³
      • A1 for 29.6% (ft)

    4. (d) The liquid is poured into a cylinder with radius 5 cm. Find the depth of the liquid in the cylinder. [2]

      • M1 for \(\pi(5)^2 h = 168\)
      • A1 for 2.13 cm

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

      • M1 for \(\frac{2}{3}\pi r^3 = 168\)
      • A1 for 4.31 cm

jmaths.xyz · AI HL · 12.10.16 Unit review