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IB Mathematics: Applications and Interpretation · Geometry and Trigonometry

Right-angled trig; sine rule; cosine rule; area of triangle practice questions.

36 exam-style questions on Right-angled trig; sine rule; cosine rule; area of triangle — topic 3.2 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
3.2
Subject
IB Mathematics: Applications and Interpretation
Topic
3 · Geometry and Trigonometry
Questions
36 with worked answers
Question types
14 fill-in-the-gap questions · 11 multiple-choice questions · 5 true/false questions · 4 flashcards · 1 diagram task · 1 matching task
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Sample questions on Right-angled trig; sine rule; cosine rule; area of triangle

In a right-angled triangle the side opposite an angle θ measures 5 cm and the hypotenuse measures 13 cm. State the value of sin θ.

  • 5/12
  • 12/13
  • 5/13
  • 13/5

Answer: 5/13

Sine is opposite over hypotenuse, so sin θ = 5/13. Choosing 5/12 confuses the adjacent side (found by Pythagoras) with the hypotenuse.

A triangle has sides a = 7, b = 9 and the included angle C = 40°. Calculate the length of side c to 3 s.f.

  • 5.79
  • 8.30
  • 10.6
  • 11.4

Answer: 5.79

By the cosine rule c² = 7² + 9² − 2(7)(9)cos40° = 49 + 81 − 96.52 = 33.48, so c = 5.79. Forgetting to subtract the 2ab cos C term gives √130 = 11.4.

In triangle ABC, angle A = 35°, side a = 8 cm and side b = 11 cm. Calculate angle B to the nearest degree (assume B is acute).

  • 23°
  • 38°
  • 52°
  • 62°

Answer: 52°

By the sine rule sinB = 11·sin35°/8 = 11(0.5736)/8 = 0.7887, so the acute B = sin⁻¹(0.7887) ≈ 52°.

Calculate the area of a triangle with sides 12 cm and 9 cm enclosing an angle of 55°, to 3 s.f.

  • 30.9 cm²
  • 44.2 cm²
  • 54.0 cm²
  • 88.4 cm²

Answer: 44.2 cm²

Area = ½(12)(9)sin55° = 54(0.8192) = 44.2 cm². Using cos55° instead of sin55° gives about 31 cm², close to the 30.9 distractor.

A triangle has all three sides known: a = 6, b = 7, c = 8. Determine angle A (opposite side a) to the nearest degree.

  • 47°
  • 58°
  • 78°
  • 82°

Answer: 47°

cosA = (b² + c² − a²)/(2bc) = (49 + 64 − 36)/(112) = 77/112 = 0.6875, so A = cos⁻¹(0.6875) = 46.6° ≈ 47°. Truncating to 46° instead of rounding is the usual slip.

Determine the length of side x opposite a 30° angle in a triangle where another side of length 10 cm lies opposite a 45° angle. Give x to 3 s.f.

  • 5.00 cm
  • 7.07 cm
  • 8.16 cm
  • 14.1 cm

Answer: 7.07 cm

By the sine rule x/sin30° = 10/sin45°, so x = 10·sin30°/sin45° = 10(0.5)/(0.7071) = 7.07 cm.

That was 6 of 36 Right-angled trig; sine rule; cosine rule; area of triangle questions.

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