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IB Mathematics: Applications and Interpretation · Geometry and Trigonometry

Applications of trig; elevation/depression; labelled diagrams practice questions.

36 exam-style questions on Applications of trig; elevation/depression; labelled diagrams — topic 3.3 of Geometry and Trigonometry in the IB Mathematics: Applications and Interpretation syllabus — with answers and explanations. Samples below; the full set is free with an account.

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About this topic

Syllabus code
3.3
Subject
IB Mathematics: Applications and Interpretation
Topic
3 · Geometry and Trigonometry
Questions
36 with worked answers
Question types
16 fill-in-the-gap questions · 10 multiple-choice questions · 6 true/false questions · 3 flashcards · 1 diagram task
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Sample questions on Applications of trig; elevation/depression; labelled diagrams

A surveyor stands 50 m from the base of a vertical tower and measures the angle of elevation to the top as 32°. Calculate the height of the tower to 3 s.f.

  • 26.5 m
  • 31.2 m
  • 42.4 m
  • 58.9 m

Answer: 31.2 m

height = 50·tan32° = 50(0.6249) = 31.2 m. Using sin32° instead of tan32° gives the incorrect 26.5 m.

A ramp rises 1.5 m over a horizontal run of 6 m. Calculate the angle of elevation of the ramp to the nearest degree.

  • 11°
  • 14°
  • 22°
  • 76°

Answer: 14°

tan θ = 1.5/6 = 0.25, so θ = tan⁻¹(0.25) ≈ 14°. Answer 76° is the complement, from dividing 6/1.5 by mistake.

A ship sails on a bearing of 120° for 30 km. Calculate how far east of its start point it now lies, to 3 s.f.

  • 15.0 km
  • 26.0 km
  • 22.5 km
  • 30.0 km

Answer: 26.0 km

A bearing of 120° is 30° south of due east, so the east component = 30·cos30° = 30·sin60° = 30(0.8660) = 26.0 km.

An aeroplane flies at a constant altitude of 2000 m. From a point on the ground the angle of elevation to the plane is 25°. Calculate the horizontal distance from that point to the point directly below the plane, to 3 s.f.

  • 845 m
  • 932 m
  • 4290 m
  • 4730 m

Answer: 4290 m

tan25° = 2000/distance, so distance = 2000/tan25° = 2000/0.4663 = 4290 m.

A hiker walks 5 km on a bearing of 060°, then 4 km on a bearing of 150°. The angle between the two legs of the path (the interior angle at the turning point) is 90°. Calculate the straight-line distance back to the start, to 3 s.f.

  • 3.00 km
  • 6.40 km
  • 6.71 km
  • 9.00 km

Answer: 6.40 km

The two bearings differ by 90°, so the legs are perpendicular; by Pythagoras the distance = √(5² + 4²) = √41 = 6.40 km.

A triangular plot has two sides of 80 m and 65 m meeting at an angle of 70°. Calculate the area of the plot to 3 s.f.

  • 1780 m²
  • 2440 m²
  • 2600 m²
  • 4890 m²

Answer: 2440 m²

Area = ½(80)(65)sin70° = 2600·(0.9397) = 2440 m². Omitting the sine factor gives the incorrect 2600 m².

That was 6 of 36 Applications of trig; elevation/depression; labelled diagrams questions.

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