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Trigonometry - Trigonometric Ratios (Sine, Cosine, Tangent)

Grade 8Cambridge (IGCSE)

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

🔑Concepts

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The Sine ratio (sin\\sin) connects the angle to the ratio of the side opposite to it and the hypotenuse: sin⁡(θ)=OppositeHypotenuse\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}.

Right-angled triangle labeling Opposite, Adjacent, and Hypotenuse relative to angle theta.
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The Cosine ratio (cos\\cos) defines the relationship between the angle and the ratio of the adjacent side to the hypotenuse: cos⁡(θ)=AdjacentHypotenuse\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}.

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The Tangent ratio (tan\\tan) relates the opposite side to the adjacent side: tan⁡(θ)=OppositeAdjacent\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}. This is useful when the hypotenuse is unknown.

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SOH CAH TOA is a common mnemonic to remember the ratios: Sine=Opp/Hyp, Cosine=Adj/Hyp, and Tangent=Opp/Adj.

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To find an unknown angle, use the inverse trigonometric functions: θ=sin⁡−1(ratio)\theta = \sin^{-1}(\text{ratio}), θ=cos⁡−1(ratio)\theta = \cos^{-1}(\text{ratio}), or θ=tan⁡−1(ratio)\theta = \tan^{-1}(\text{ratio}).

📐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).

Problem 4:

A ladder leans against a vertical wall. The foot of the ladder is 3 m3 \text{ m} from the wall and the ladder makes an angle of 70∘70^{\circ} with the ground. Calculate the height hh the ladder reaches up the wall.

A ladder forming a triangle with a wall and the ground.

Solution:

  1. Identify the sides relative to the 70∘70^{\circ} angle: Adjacent side =3 m= 3 \text{ m}, Opposite side =h= h.
  2. Choose the Tangent ratio: tan⁡(70∘)=OppositeAdjacent\tan(70^{\circ}) = \frac{\text{Opposite}}{\text{Adjacent}}.
  3. Substitute values: tan⁡(70∘)=h3\tan(70^{\circ}) = \frac{h}{3}.
  4. Rearrange for hh: h=3×tan⁡(70∘)h = 3 \times \tan(70^{\circ}).
  5. Calculate: h≈3×2.7475=8.2425 mh \approx 3 \times 2.7475 = 8.2425 \text{ m}.

Explanation:

Since we have the adjacent side and need the opposite side, Tangent is the correct ratio to use.

Problem 5:

Calculate the length of the hypotenuse xx in a right-angled triangle where the side adjacent to a 40∘40^{\circ} angle is 12 cm12 \text{ cm}.

Right-angled triangle with base 12 cm, angle 40 degrees and hypotenuse x.

Solution:

  1. Identify the sides: Adjacent side =12 cm= 12 \text{ cm}, Hypotenuse =x= x.
  2. Choose the Cosine ratio: cos⁡(40∘)=AdjacentHypotenuse\cos(40^{\circ}) = \frac{\text{Adjacent}}{\text{Hypotenuse}}.
  3. Substitute values: cos⁡(40∘)=12x\cos(40^{\circ}) = \frac{12}{x}.
  4. Rearrange for xx: x=12cos⁡(40∘)x = \frac{12}{\cos(40^{\circ})}.
  5. Calculate: x≈120.7660=15.665 cmx \approx \frac{12}{0.7660} = 15.665 \text{ cm}.

Explanation:

We use Cosine because the information given includes the adjacent side and requires the hypotenuse.