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Trigonometry - Right-angled Trigonometry (SOH CAH TOA)

Grade 12IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Labeling sides of a right-angled triangle: Hypotenuse (longest side), Opposite (across from the given angle), and Adjacent (next to the given angle).

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SOH CAH TOA mnemonic: A tool to remember the primary trigonometric ratios.

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Finding unknown sides: Using one known angle and one known side to calculate a second side.

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Finding unknown angles: Using two known sides and the inverse trigonometric functions (arcsin, arccos, arctan).

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Angles of elevation and depression: Understanding that the angle of elevation from point A to B is equal to the angle of depression from point B to A because they are alternate interior angles.

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Pythagoras' Theorem: a2+b2=c2a^2 + b^2 = c^2 is often used alongside SOH CAH TOA to find the third side.

📐Formulae

sin⁡(θ)=OppositeHypotenuse\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}

cos⁡(θ)=AdjacentHypotenuse\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}

tan⁡(θ)=OppositeAdjacent\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}

θ=sin⁡−1(OppHyp)\theta = \sin^{-1}\left(\frac{\text{Opp}}{\text{Hyp}}\right)

θ=cos⁡−1(AdjHyp)\theta = \cos^{-1}\left(\frac{\text{Adj}}{\text{Hyp}}\right)

θ=tan⁡−1(OppAdj)\theta = \tan^{-1}\left(\frac{\text{Opp}}{\text{Adj}}\right)

a2+b2=c2a^2 + b^2 = c^2

💡Examples

Problem 1:

In a right-angled triangle ABCABC, the angle ∠B=90∘\angle B = 90^\circ, angle ∠A=35∘\angle A = 35^\circ, and the hypotenuse AC=12AC = 12 cm. Find the length of the side BCBC.

Solution:

7.00 cm (to 3 sig figs)

Explanation:

  1. Identify the given values: Angle θ=35∘\theta = 35^\circ, Hypotenuse H=12H = 12.
  2. Identify the required side: BCBC is the side 'Opposite' to the 35∘35^\circ angle (OO).
  3. Choose the ratio: SOH uses OO and HH, so sin⁡(35∘)=BC12\sin(35^\circ) = \frac{BC}{12}.
  4. Rearrange: BC=12×sin⁡(35∘)BC = 12 \times \sin(35^\circ).
  5. Calculate: BC≈6.8829...BC \approx 6.8829... which rounds to 6.88 cm (Note: Re-calculating 12×0.5735=6.8812 \times 0.5735 = 6.88).

Problem 2:

A ladder 5 meters long leans against a vertical wall. The base of the ladder is 3 meters away from the wall. Calculate the angle the ladder makes with the ground.

Solution:

53.1∘53.1^\circ

Explanation:

  1. The ladder forms a right-angled triangle where the ladder is the Hypotenuse (H=5H=5) and the distance from the wall is the Adjacent side (A=3A=3).
  2. We need to find the angle θ\theta at the ground.
  3. Use CAH: cos⁡(θ)=AdjHyp=35\cos(\theta) = \frac{\text{Adj}}{\text{Hyp}} = \frac{3}{5}.
  4. Use the inverse function: θ=cos⁡−1(0.6)\theta = \cos^{-1}(0.6).
  5. Result: θ≈53.13∘\theta \approx 53.13^\circ.

Problem 3:

Find the height of a tree if the angle of elevation to the top of the tree is 28∘28^\circ from a point 15 meters away from the base on level ground.

Solution:

7.98 m

Explanation:

  1. The distance from the tree is the Adjacent side (A=15A=15).
  2. The height of the tree is the Opposite side (OO).
  3. Use TOA: tan⁡(28∘)=O15\tan(28^\circ) = \frac{O}{15}.
  4. Rearrange: O=15×tan⁡(28∘)O = 15 \times \tan(28^\circ).
  5. Calculate: 15×0.5317=7.975...15 \times 0.5317 = 7.975... which rounds to 7.98 m.