Right-Angle Trigonometry (SOHCAHTOA)

\( \sin\theta=\tfrac{\text{opp}}{\text{hyp}},\quad \cos\theta=\tfrac{\text{adj}}{\text{hyp}},\quad \tan\theta=\tfrac{\text{opp}}{\text{adj}} \)
Trigonometry GCSE
Question 8 of 20
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A kite string is 50m long and makes an angle of 70° with the ground. Find the height of the kite (2dp).

Tips: use ^ for powers, sqrt() for roots, and type pi for π.
Hint (H)
\( Use sin θ = opp/hyp \)

Explanation

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Statement

In a right-angled triangle, the trigonometric ratios are defined as:

\[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}, \quad \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}, \quad \tan \theta = \frac{\text{opposite}}{\text{adjacent}}. \]

Why it’s true

  • These ratios come from comparing side lengths in right-angled triangles.
  • The hypotenuse is always the longest side (opposite the right angle).
  • The “opposite” and “adjacent” sides are defined relative to the chosen angle \(\theta\).
  • The definitions work for all right-angled triangles because they scale proportionally (similar triangles property).

Recipe (how to use it)

  1. Label the triangle: hypotenuse, opposite, adjacent (relative to \(\theta\)).
  2. Select SOH (sin), CAH (cos), or TOA (tan) depending on which sides are involved.
  3. Write down the equation.
  4. Rearrange to find the unknown side or angle.

Spotting it

Whenever a right-angled triangle involves a missing side or angle and you know another side/angle, SOHCAHTOA applies.

Common pairings

  • Finding side lengths in right triangles.
  • Finding angles from side lengths.
  • Applications in heights, distances, and ladders against walls.

Mini examples

  1. Triangle with hypotenuse 10, angle \(\theta=30^\circ\). Opposite = \(10\sin 30^\circ = 5\).
  2. Triangle with adjacent=8, hypotenuse=10. \(\cos \theta = 8/10 = 0.8 → \theta≈36.87^\circ\).
  3. Triangle with opposite=7, adjacent=24. \(\tan \theta = 7/24 → \theta≈16.26^\circ\).

Pitfalls

  • Choosing the wrong side as opposite/adjacent.
  • Forgetting to check calculator is in degree mode.
  • Mixing up sine, cosine, and tangent ratios.

Exam strategy

  • Draw and label the triangle first.
  • Pick the correct SOH/CAH/TOA formula.
  • Rearrange carefully using inverse trig for angles.
  • Round answers to 1–2 decimal places unless exact values are required.

Summary

SOHCAHTOA is the foundation of trigonometry in right triangles. It provides quick, consistent methods to calculate missing sides or angles in applied problems.

Worked examples

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  1. \( In a right triangle with hypotenuse 10 and angle θ=30°, find the opposite side. \)
    1. \( sin θ = opp/hyp \)
    2. \( sin 30°=opp/10 \)
    3. \( 0.5=opp/10 → opp=5 \)
    Answer: 5
  2. \( In a right triangle, adjacent=8 and hypotenuse=10. Find angle θ. \)
    1. \( cos θ = adj/hyp \)
    2. \( cos θ=8/10=0.8 \)
    3. \( θ=cos⁻¹(0.8)=36.87° \)
    Answer: 36.9° (approx)
  3. \( In a right triangle, opposite=6 and adjacent=8. Find tan θ. \)
    1. \( tan θ=opp/adj \)
    2. \( tan θ=6/8=0.75 \)
    Answer: 0.75
  4. A ladder 5m long rests against a wall, making an angle of 60° with the ground. Find the height it reaches.
    1. \( sin θ = opp/hyp \)
    2. \( sin 60°=opp/5 \)
    3. \( 0.866=opp/5 → opp=4.33m \)
    Answer: 4.33m
  5. \( In a right triangle with opposite=7 and adjacent=24, find θ. \)
    1. \( tan θ=opp/adj \)
    2. \( tan θ=7/24 \)
    3. \( θ=tan⁻¹(7/24)≈16.26° \)
    Answer: 16.3° (approx)
  6. \( In a right triangle, θ=45°, hypotenuse=12. Find adjacent side. \)
    1. \( cos θ=adj/hyp \)
    2. \( cos 45°=adj/12 \)
    3. \( 0.707=adj/12 → adj≈8.49 \)
    Answer: 8.49
  7. A building casts a 20m shadow. The angle of elevation of the sun is 30°. Find the building’s height.
    1. \( tan θ=opp/adj \)
    2. \( tan 30°=opp/20 \)
    3. \( 0.577=opp/20 → opp≈11.55m \)
    Answer: 11.6m (approx)
  8. \( In a right triangle, opposite=9, hypotenuse=15. Find angle θ. \)
    1. \( sin θ=opp/hyp \)
    2. \( sin θ=9/15=0.6 \)
    3. \( θ=sin⁻¹(0.6)≈36.87° \)
    Answer: 36.9° (approx)
  9. A ramp of length 6m rises to a platform 1.5m high. Find the angle of elevation θ.
    1. \( sin θ=opp/hyp \)
    2. \( sin θ=1.5/6=0.25 \)
    3. \( θ=sin⁻¹(0.25)≈14.48° \)
    Answer: 14.5° (approx)
  10. \( In a right triangle, adjacent=5, opposite=12. Find hypotenuse. \)
    1. Use Pythagoras or trig
    2. \( tan θ=12/5 but easier: hyp=√(5²+12²)=13 \)
    Answer: 13