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MPM2D Trigonometry
Grade 10 Unit 6 — practice questions with full solutions

Primary trig ratios, similar triangles, the sine law and the cosine law — including how to tell which one a question actually needs.

Free MPM2D trigonometry practice for Grade 10 students in Ontario. These 6 questions cover the same material as a typical unit test or exam review on this topic, and every one comes with a complete worked solution — not just an answer key. Useful whether you are preparing for a MPM2D unit test, catching up on a lesson, or reviewing before the final exam.
MPM2D · Grade 10 Trigonometry 5 core 1 challenge
1Core
12 cmb = ?a = ?35°ABC
For the right triangle shown, find sides a and b to one decimal place.
Show the full solution
  1. Angle B is 35° and the hypotenuse is 12 cm.
  2. Side b (from A to C) is opposite the 35° angle, so use sine: sin 35° = b ÷ 12.
  3. b = 12 sin 35° ≈ 12 × 0.5736 ≈ 6.9 cm.
  4. Side a (from A to B) is adjacent to 35°, so use cosine: a = 12 cos 35° ≈ 12 × 0.8192 ≈ 9.8 cm.
  5. Check with Pythagoras: 6.9² + 9.8² ≈ 47.6 + 96.0 ≈ 143.6, and 12² = 144. ✔
Final answera ≈ 9.8 cm, b ≈ 6.9 cm
2Core
a = 8 cmb = ?40°75°ABC
In the triangle shown, angle A = 40°, angle B = 75° and side a = 8 cm. Find side b to one decimal place, and state angle C.
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  1. Angles in a triangle sum to 180°, so C = 180° − 40° − 75° = 65°.
  2. Two angles and a side opposite one of them means the sine law applies.
  3. a ÷ sin A = b ÷ sin B, so b = a sin B ÷ sin A.
  4. b = 8 × sin 75° ÷ sin 40° ≈ 8 × 0.9659 ÷ 0.6428.
  5. That gives b ≈ 12.0 cm.
Final answerC = 65°, b ≈ 12.0 cm
3Core
9 cm7 cmc = ?52°ABC
In the triangle shown, two sides are 7 cm and 9 cm with an included angle of 52°. Find the third side c to one decimal place, and explain why the sine law cannot be used first here.
Show the full solution
  1. You have two sides and the angle between them, which is the cosine law situation.
  2. The sine law needs a known angle opposite a known side, and here the 52° angle is between the two given sides — no such pair exists yet.
  3. c² = a² + b² − 2ab cos C = 7² + 9² − 2(7)(9) cos 52°.
  4. Calculate: 49 + 81 − 126 × 0.61566 = 130 − 77.57 = 52.43.
  5. c = √52.43 ≈ 7.2 cm.
Final answerc ≈ 7.2 cm; the cosine law is needed because the known angle sits between the two known sides
4Core
A 6 m ladder leans against a wall, reaching 5.2 m up. Find the angle the ladder makes with the ground, to the nearest degree.
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  1. The wall height is opposite the angle and the ladder is the hypotenuse, so use sine.
  2. sin θ = 5.2 ÷ 6 = 0.8667.
  3. Take the inverse sine: θ = sin⁻¹(0.8667).
  4. That gives θ ≈ 60.07°.
Final answer≈ 60°
5Core
Two triangles are similar. The first has sides 6, 8 and 10. The longest side of the second is 25. Find its other two sides and the ratio of their areas.
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  1. Similar triangles have proportional sides. The scale factor is 25 ÷ 10 = 2.5.
  2. Multiply the other sides: 6 × 2.5 = 15 and 8 × 2.5 = 20.
  3. Areas of similar figures scale by the square of the side ratio.
  4. So the area ratio is 2.5² = 6.25, or 25 : 4 when written with the smaller first as 1 : 6.25.
Final answersides 15 and 20; areas in the ratio 1 : 6.25
6Challenge
From a point 40 m from the base of a tower, the angle of elevation to the top is 38°. From a second point further away in the same direction, the angle is 22°. How far apart are the two points, to one decimal place?
Show the full solution
  1. Let the tower height be h. From the first point: tan 38° = h ÷ 40, so h = 40 tan 38° ≈ 40 × 0.78129 ≈ 31.25 m.
  2. From the second point at distance d: tan 22° = h ÷ d, so d = h ÷ tan 22°.
  3. d ≈ 31.25 ÷ 0.40403 ≈ 77.35 m.
  4. The separation is d − 40 ≈ 77.35 − 40 = 37.35 m.
Final answer≈ 37.4 m apart

MPM2D Trigonometry — common questions

Short answers to the things students ask most about this unit.

How do I know whether to use the sine law or the cosine law?

Use the sine law when you have a matching angle-and-opposite-side pair. Use the cosine law for two sides with the angle between them, or for all three sides.

What is SOH CAH TOA used for?

It is the memory aid for the three primary ratios in a right triangle only: sine is opposite over hypotenuse, cosine adjacent over hypotenuse, tangent opposite over adjacent.

More MPM2D practice — grade 10 academic

MPM2D Linear SystemsMPM2D Analytic GeometryMPM2D Factoring QuadraticsMPM2D Quadratic Relations & GraphingAll MPM2D units
← Unit 5: Solving Quadratic EquationsBack to all assignments →
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