L50 — Geometry, Measurement and Trigonometry Review — Solutions

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1. Q1 — Right-triangle trigonometry

   Fluency
   • (a) AC = √(AB²−BC²) = √(169−25) = √144 = 12 m.
   • (b) sin A = 5/13, cos A = 12/13, tan A = 5/12.
   • (c) A = sin⁻¹(5/13) ≈ 22.6°.
   • (d) B = 90° − 22.6° = 67.4°.

2. Q2 — Sine and cosine rules

   Fluency
   • (a) Angle R = 180°−50°−70° = 60°. QR/sin P = PQ/sin R ⇒ QR = 12 sin 50°/sin 60° ≈ 12(0.766)/0.866 ≈ 10.6 cm.
   • (b) XZ² = 49+81−2(7)(9)cos55° ≈ 130−126(0.574) ≈ 130−72.3 = 57.7. XZ ≈ 7.6 cm.
   • (c) A = ½(7)(9)sin55° ≈ ½(63)(0.819) ≈ 25.8 cm².
   • (d) cos E = (DE²+EF²−DF²)/(2×DE×EF) = (25+64−100)/80 = −11/80 ≈ −0.1375. E = cos⁻¹(−0.1375) ≈ 97.9°.

3. Q3 — Surface area and volume

   Fluency
   • (a) SA = 2π(9)+2π(3)(8) = 18π+48π = 66π cm².
   • (b) V = ⅓π(36)(10) = 120π cm³.
   • (c) r=7. SA = 4π(49) = 196π ≈ 615.8 cm².
   • (d) Base = 100 m². 4 triangular faces: 4 × ½(10)(13) = 260 m². Total SA = 100+260 = 360 m².

4. Q4 — Circle geometry theorems

   Fluency
   • (a) Angle at circumference = ½ × 140° = 70°.
   • (b) Opposite angles sum to 180°: R = 180°−105° = 75°. S = 180°−80° = 100°.
   • (c) TA = √(OT²−OA²) = √(169−25) = √144 = 12 cm.
   • (d) Tangents from an external point are equal: PB = PA = 8 cm.

5. Q5 — Area of a triangle and applications

   Understanding
   • (a) A = ½(80)(60)sin72° ≈ ½(4800)(0.951) ≈ 2282 m².
   • (b) c² = 6400+3600−2(80)(60)cos72° ≈ 10000−9600(0.309) ≈ 10000−2966 = 7034. c ≈ 83.9 m.
   • (c) Cost = 83.9 × 45 ≈ $3 776.
   • (d) sin A/BC = sin C/AB (using sine rule with known A=72° side opposite, etc.) sin A = 60 sin72°/83.9 ≈ 60(0.951)/83.9 ≈ 0.680. A ≈ 42.8°.

6. Q6 — Composite solids

   Understanding
   • (a) V = π(4)(5) + ½×&frac43;π(8) = 20π + 16π/3 = (60π+16π)/3 = 76π/3 m³.
   • (b) Base circle (πr²=4π) + cylinder side (2πrh=20π) + hemisphere curved surface (2πr²=8π) = 32π m².
   • (c) V = 76π/3 ≈ 76(3.14)/3 ≈ 79.5 m³. Time = 79.5/0.5 = 159 minutes.
   • (d) Hemisphere height = radius = 2 m. Volume of hemisphere ≈ 16π/3 ≈ 16.8 m³. Time to drain hemisphere = 16.8/(0.2×πr²) ... rate of level drop = 0.2 m/h. The hemisphere drains first; its 2 m depth takes 2/0.2 = 10 hours.

7. Q7 — Angles of elevation and bearings

   Understanding
   • (a) h = 80 tan 32° ≈ 80 × 0.625 ≈ 50 m.
   • (b) Angle between AB (bearing 060°) and BC (bearing 150°) is 150°−60°=90°. AC²=20²+15²=625. AC=25 km.
   • (c) tan(bearing angle from A)=15/20 ⇒ θ=36.9° from AB direction. Bearing of C from A = 060°+36.9°≈097°.
   • (d) tanθ=3000/(5000)=0.6. θ=tan⁻¹(0.6)≈31.0°.

8. Q8 — Similarity and scale factors

   Understanding
   • (a) Scale factor = 10/4 = 5/2.
   • (b) SA ratio = (5/2)² = 25/4.
   • (c) Volume ratio = (5/2)³ = 125/8.
   • (d) V_large = 64π × 125/8 = 1 000π cm³.

9. Q9 — Navigation problem

   Problem Solving
   • (a) Ship A: 30×2=60 km. Ship B: 25×2=50 km.
   • (b) Bearing of A: 040°. Bearing of B: 310°. Angle between paths = 360°−310°+40° = 90° ... More carefully: from north clockwise 040° and 310°. Angle between = 310°−040°=270° going one way, or 90° going the other. Interior angle between the two directions = 360°−270° = 90°.
   • (c) d²=60²+50²−2(60)(50)cos90°=3600+2500−0=6100. d=√6100≈78.1 km.
   • (d) Bearing of B from A: the perpendicular arrangement means A is northeast and B is northwest of P. From A's position (NE of P), B is to the west-northwest. Using tan of the right triangle: tanθ=50/60. θ≈39.8° west of south from A. Bearing ≈ 270° − (90°−39.8°) = 219.8° ≈ 220°.

10. Q10 — Extended: geometry and measurement

    Problem Solving
    • (a) r = √(l²−h²) = √(169−144) = √25 = 5 cm.
    • (b) V = ⅓π(25)(12) = 100π cm³.
    • (c) &frac43;πR³ = 100π ⇒ R³ = 75 ⇒ R = ∛75 = 5∛3 × ∛(3/5) ... = (75)^1/3 ≈ 4.22 cm.
    • (d) Cone SA = πr²+πrl = π(25)+π(5)(13) = 25π+65π = 90π ≈ 282.6 cm². Sphere SA = 4πR² = 4π(75)^2/3 ≈ 4π(17.80) ≈ 223.5 cm². The cone has greater surface area (282.6 > 223.5 cm²).