Every right-angled trig question in Grade 10 — naming sides, finding a side with sin/cos/tan, finding the hypotenuse, finding an angle with the inverse, special angles, and an angle-of-elevation application — each worked in full, with a diagram and worksheet.
Trigonometry connects the angles of a right-angled triangle to the lengths of its sides. It starts with three ratios — sine, cosine and tangent — and once you can name the sides correctly, every question type follows the same routine.
1Name the three sides
- The hypotenuse is the longest side, always opposite the right angle.
- The opposite side is across from the angle you're working with.
- The adjacent side is next to that angle (but is not the hypotenuse).
Important: 'opposite' and 'adjacent' depend on which angle you're using. Name them fresh for each question.
2The memory trick: SOH-CAH-TOA
The routine every time: (1) label the sides, (2) pick the ratio that uses the two sides in play, (3) solve.
3Type 1: Find a side using sine (opposite & hypotenuse)
- 1You have the hypotenuse and want the opposite → use sine (SOH).Sine links opposite and hypotenuse.
- 2sin 30° = x / 10.Opposite over hypotenuse.
- 3x = 10 × sin 30° = 10 × 0,5 = 5.Multiply both sides by 10.
4Type 2: Find a side using cosine or tangent
- 1Adjacent and hypotenuse → use cosine (CAH).cos = adj/hyp.
- 2cos 40° = x / 12 → x = 12 × cos 40°.Isolate x.
- 3x ≈ 12 × 0,766 = 9,19.Use the calculator (in degrees).
- 1Opposite and adjacent → use tangent (TOA).tan = opp/adj.
- 2tan 35° = x / 8 → x = 8 × tan 35°.
- 3x ≈ 8 × 0,700 = 5,60.
5Type 3: Find the hypotenuse
- 1Opposite and hypotenuse → sine: sin 25° = 6 / h.The unknown is on the bottom this time.
- 2Rearrange: h = 6 / sin 25°.Multiply both sides by h, divide by sin 25°.
- 3h ≈ 6 / 0,4226 = 14,2.
6Type 4: Find an angle (use the inverse)
When you know two sides and want the angle, use the inverse function (sin−1, cos−1, tan−1 on your calculator).
- 1Opposite and hypotenuse → sine: sin θ = 710 = 0,7.Set up the ratio first.
- 2θ = sin−1(0,7).Inverse sine 'undoes' the sine to give the angle.
- 3θ ≈ 44,4°.Use the sin−1 (or 2ndF sin) key.
7Type 5: Special angles (exact values)
Memorise these — they come up without a calculator: sin 30° = 12, cos 60° = 12, sin 60° = cos 30° = 32, sin 45° = cos 45° = 12, tan 45° = 1.
8Type 6: Application: angle of elevation
- 1The 4 m is opposite the 60° angle; the ladder is the hypotenuse → sine.opp and hyp → SOH.
- 2sin 60° = 4 / L → L = 4 / sin 60°.The unknown hypotenuse is on the bottom.
- 3L ≈ 4 / 0,866 = 4,62 m.
Finding a side? Use sin/cos/tan directly. Finding an angle? Use the inverse (sin−1 etc). And always keep your calculator in DEGREES — wrong mode is the #1 cause of 'random' wrong answers.
Pick the ratio from the two sides you have. Don't default to sine — if you have opposite and adjacent, it's tangent. Label the sides first, every time.
Trig is a big, reliable section of Paper 2 in Grades 11–12. Master these six question types now and the harder work later builds on solid ground.
Practice exercises
Work each one out, then click to reveal the answer.
- 1Find x opposite a 30° angle, hypotenuse 8. (sine)
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x = 8 sin30° = 8×0.5 = 4 - 2Find x opposite a 50° angle, hypotenuse 10. (sine)
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10 sin50° ≈ 7.66 - 3Find x adjacent to a 60° angle, hypotenuse 14. (cosine)
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14 cos60° = 14×0.5 = 7 - 4Find x opposite a 40° angle, adjacent 9. (tangent)
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9 tan40° ≈ 7.55 - 5The side opposite a 30° angle is 5. Find the hypotenuse. (hypotenuse)
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h = 5/sin30° = 10 - 6opposite = 6, hypotenuse = 12. Find θ. (angle)
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sinθ = 0.5 → θ = 30° - 7adjacent = 8, hypotenuse = 16. Find θ. (angle)
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cosθ = 0.5 → θ = 60° - 8opposite = 5, adjacent = 5. Find θ. (angle)
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tanθ = 1 → θ = 45° - 9Give the exact value of sin 30° and tan 45°. (special)
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12 and 1 - 10A rope to the top of a 6 m pole makes 40° with the ground. Rope length? (application)
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sin40° = 6L → L = 6/sin40° ≈ 9,33 m
Now practise it
Download Grade 10 past papers and worksheets on this topic.