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Trigonometry - Trigonometric Ratios (Sine, Cosine, Tangent)

Grade 8IGCSE

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

🔑Concepts

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Right-Angled Triangles: Trigonometric ratios only apply to triangles with a 90-degree angle at this level.

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Side Identification: Sides are labeled based on their position relative to the reference angle (theta): Hypotenuse (longest side), Opposite (across from the angle), and Adjacent (next to the angle).

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SOH CAH TOA: A mnemonic used to remember which sides relate to which trigonometric function.

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Constant Ratios: For a fixed angle, the ratio of the sides remains constant regardless of the triangle's size.

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Inverse Functions: Used to find a missing angle when two side lengths are already known (sin⁻¹, cos⁻¹, tan⁻¹).

📐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=tan⁡−1(OppositeAdjacent)\theta = \tan^{-1}\left(\frac{\text{Opposite}}{\text{Adjacent}}\right)

💡Examples

Problem 1:

In a right-angled triangle, the hypotenuse is 10 cm and the angle θ\theta is 30°. Calculate the length of the opposite side.

Solution:

Opposite = 10×sin⁡(30∘)=10×0.5=510 \times \sin(30^\circ) = 10 \times 0.5 = 5 cm

Explanation:

We are given the Hypotenuse (10) and an Angle (30°), and we need to find the Opposite side. According to SOH, we use the Sine ratio. Rearranging sin⁡(30)=O10\sin(30) = \frac{O}{10} gives O=10×sin⁡(30)O = 10 \times \sin(30).

Problem 2:

A right-angled triangle has an adjacent side of 7 cm and an opposite side of 5 cm. Find the value of the angle θ\theta.

Solution:

θ=tan⁡−1(57)≈35.54∘\theta = \tan^{-1}(\frac{5}{7}) \approx 35.54^\circ

Explanation:

We are given the Opposite (5) and Adjacent (7) sides. According to TOA, we use the Tangent ratio. To find the angle, we use the inverse tangent function: θ=tan⁡−1(57)\theta = \tan^{-1}(\frac{5}{7}).

Problem 3:

Find the length of the adjacent side if the hypotenuse is 15 cm and the angle is 60°.

Solution:

Adjacent = 15×cos⁡(60∘)=15×0.5=7.515 \times \cos(60^\circ) = 15 \times 0.5 = 7.5 cm

Explanation:

We have the Hypotenuse and need the Adjacent side. According to CAH, we use the Cosine ratio. Rearranging cos⁡(60)=A15\cos(60) = \frac{A}{15} gives A=15×cos⁡(60)A = 15 \times \cos(60).