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Trigonometry - Right-angled Trigonometry (SOHCAHTOA)

Grade 9IGCSE

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

🔑Concepts

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

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The SOHCAHTOA mnemonic: Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent.

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Using trigonometric ratios to find missing side lengths when one angle and one side are known.

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Using inverse trigonometric functions (sin⁻¹, cos⁻¹, tan⁻¹) to find missing angles when two sides are known.

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The requirement of ensuring the calculator is set to 'DEG' (Degrees) mode for IGCSE calculations.

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Rounding rules: Generally, IGCSE requires angles to be rounded to 1 decimal place and side lengths to 3 significant figures unless stated otherwise.

📐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)

💡Examples

Problem 1:

In a right-angled triangle, the angle θ\theta is 32∘32^\circ and the hypotenuse is 1515 cm. Find the length of the opposite side xx.

Solution:

x=15×sin⁡(32∘)≈7.95x = 15 \times \sin(32^\circ) \approx 7.95 cm

Explanation:

Identify the knowns: Hypotenuse = 15, Angle = 32°. We need the Opposite side. SOH tells us to use Sine. sin⁡(32∘)=x15\sin(32^\circ) = \frac{x}{15}. Multiplying both sides by 15 gives x=15sin⁡(32∘)x = 15 \sin(32^\circ).

Problem 2:

A ladder 55 m long leans against a vertical wall. The base of the ladder is 33 m away from the wall. Calculate the angle the ladder makes with the ground.

Solution:

θ=cos⁡−1(35)=53.1∘\theta = \cos^{-1}(\frac{3}{5}) = 53.1^\circ

Explanation:

The ladder represents the Hypotenuse (5m) and the distance from the wall is the Adjacent side (3m). CAH tells us to use Cosine: cos⁡(θ)=35\cos(\theta) = \frac{3}{5}. To find the angle, use the inverse cosine function.

Problem 3:

Find the length of the adjacent side if the opposite side is 88 cm and the angle is 40∘40^\circ.

Solution:

Adjacent=8tan⁡(40∘)≈9.53\text{Adjacent} = \frac{8}{\tan(40^\circ)} \approx 9.53 cm

Explanation:

We have the opposite side and need the adjacent side. TOA tells us to use Tangent. tan⁡(40∘)=8adj\tan(40^\circ) = \frac{8}{\text{adj}}. Rearranging for 'adj' gives adj=8tan⁡(40∘)\text{adj} = \frac{8}{\tan(40^\circ)}.