krit.club logo

Trigonometry - Right-angled Trigonometry (SOH CAH TOA)

Grade 12A Level

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

🔑Concepts

•

The naming of sides in a right-angled triangle depends on the position of the reference angle θ\theta. The 'Hypotenuse' is always the longest side opposite the 90∘90^\circ angle, the 'Opposite' side is across from θ\theta, and the 'Adjacent' side is next to θ\theta.

A right-angled triangle showing the naming of sides relative to angle theta.
•

The mnemonic SOH CAH TOA is used to remember the three primary trigonometric ratios: sin⁡(θ)=OH\sin(\theta) = \frac{O}{H}, cos⁡(θ)=AH\cos(\theta) = \frac{A}{H}, and tan⁡(θ)=OA\tan(\theta) = \frac{O}{A}.

•

To find an unknown angle when two sides are known, use the inverse trigonometric functions: sin⁡−1\sin^{-1}, cos⁡−1\cos^{-1}, or tan⁡−1\tan^{-1}.

•

The Angle of Elevation is the angle measured upwards from the horizontal line of sight to an object above. Conversely, the Angle of Depression is the angle measured downwards from the horizontal to an object below.

Diagram showing the angle of elevation from a horizontal line to an object.

📐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.

Problem 4:

A surveyor stands 4040 m away from the base of a building and measures the angle of elevation to the top of the building as 38∘38^\circ. Determine the height of the building to two decimal places.

A right-angled triangle representing a building height problem.

Solution:

  1. Identify knowns: Angle θ=38∘\text{Angle } \theta = 38^\circ, Adjacent side =40\text{Adjacent side } = 40 m.
  2. Identify unknown: Opposite side (h)\text{Opposite side } (h).
  3. Use the Tangent ratio: tan⁡(38∘)=h40\tan(38^\circ) = \frac{h}{40}.
  4. Solve for hh: h=40×tan⁡(38∘)h = 40 \times \tan(38^\circ).
  5. h≈40×0.7813=31.2515h \approx 40 \times 0.7813 = 31.2515.
  6. The height is 31.2531.25 m.

Explanation:

Since we have the distance from the base (Adjacent) and want to find the height (Opposite), we use TOA (tan=O/A\\tan = O/A).

Problem 5:

A slide is 6.56.5 m long. Its vertical height is 3.23.2 m. Calculate the angle xx that the slide makes with the horizontal ground.

A right-angled triangle representing a slide of 6.5m and height 3.2m.

Solution:

  1. Identify knowns: Hypotenuse =6.5\text{Hypotenuse } = 6.5 m, Opposite side =3.2\text{Opposite side } = 3.2 m.
  2. Identify unknown: Angle x\text{Angle } x.
  3. Use the Sine ratio: sin⁡(x)=3.26.5\sin(x) = \frac{3.2}{6.5}.
  4. Solve for xx: x=sin⁡−1(3.26.5)x = \sin^{-1}\left(\frac{3.2}{6.5}\right).
  5. x≈sin⁡−1(0.4923)x \approx \sin^{-1}(0.4923).
  6. x≈29.5∘x \approx 29.5^\circ.

Explanation:

We use the SOH ratio (Sine = Opposite / Hypotenuse) because we are given the length of the slide (Hypotenuse) and its height (Opposite).