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Trigonometry - Calculating Sides and Angles in Right-Angled Triangles

Grade 8Cambridge (IGCSE)

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

🔑Concepts

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In a right-angled triangle, the three sides are named relative to a specific angle θ\theta: the Hypotenuse (opposite the 90∘90^{\circ} angle), the Opposite (across from θ\theta), and the Adjacent (next to θ\theta).

A right-angled triangle labeling the opposite, adjacent, and hypotenuse sides relative to an angle theta.
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The SOH CAH TOA mnemonic helps remember the 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}.

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To find an unknown side, identify the two sides involved (one known, one unknown) and the given angle, then choose the ratio that links them.

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

📐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}}

a2+b2=c2a^2 + b^2 = c^2 (Pythagoras' Theorem)

θ=sin⁡−1(OppositeHypotenuse)\theta = \sin^{-1}\left(\frac{\text{Opposite}}{\text{Hypotenuse}}\right)

💡Examples

Problem 1:

In a right-angled triangle, the hypotenuse is 12 cm and one angle is 35°. Calculate the length of the side opposite to the 35° angle. Give your answer to 2 decimal places.

Solution:

6.88 cm

Explanation:

  1. Identify the given information: Angle θ=35∘\theta = 35^\circ, Hypotenuse = 12 cm. We need the Opposite side. 2. Choose the ratio: SOH uses Opposite and Hypotenuse. 3. Set up the equation: sin⁡(35∘)=x12\sin(35^\circ) = \frac{x}{12}. 4. Solve for xx: x=12×sin⁡(35∘)≈12×0.573576=6.8829...x = 12 \times \sin(35^\circ) \approx 12 \times 0.573576 = 6.8829... 5. Round to 2 decimal places.

Problem 2:

A right-angled triangle has an adjacent side of 7 cm and an opposite side of 5 cm relative to an angle θ\theta. Find the value of θ\theta to 1 decimal place.

Solution:

35.5°

Explanation:

  1. Identify the given information: Opposite = 5 cm, Adjacent = 7 cm. 2. Choose the ratio: TOA uses Opposite and Adjacent. 3. Set up the equation: tan⁡(θ)=57\tan(\theta) = \frac{5}{7}. 4. Use the inverse tangent function: θ=tan⁡−1(57)\theta = \tan^{-1}(\frac{5}{7}). 5. Calculate: θ≈tan⁡−1(0.7142)=35.537...\theta \approx \tan^{-1}(0.7142) = 35.537... 6. Round to 1 decimal place.

Problem 3:

A ladder 5 m long leans against a vertical wall. The base of the ladder is 3 m away from the wall. Calculate the angle α\alpha that the ladder makes with the ground. Give your answer to 1 decimal place.

A ladder leaning against a wall forming a right-angled triangle with hypotenuse 5 and base 3.

Solution:

cos⁡(α)=AdjacentHypotenuse\cos(\alpha) = \frac{\text{Adjacent}}{\text{Hypotenuse}} cos⁡(α)=35\cos(\alpha) = \frac{3}{5} α=cos⁡−1(0.6)\alpha = \cos^{-1}(0.6) α≈53.1∘\alpha \approx 53.1^{\circ}

Explanation:

Identify the sides: the ladder is the hypotenuse (55 m) and the distance from the wall is the adjacent side (33 m) to the angle at the ground. Using the cosine ratio allows us to solve for the angle.

Problem 4:

In triangle XYZXYZ, angle Y=90∘Y = 90^{\circ}, angle Z=42∘Z = 42^{\circ} and the side XY=8XY = 8 cm. Calculate the length of the adjacent side YZYZ. Give your answer to 2 decimal places.

Right-angled triangle XYZ with angle Y=90, angle Z=42 and XY=8cm.

Solution:

tan⁡(42∘)=OppositeAdjacent\tan(42^{\circ}) = \frac{\text{Opposite}}{\text{Adjacent}} tan⁡(42∘)=8YZ\tan(42^{\circ}) = \frac{8}{YZ} YZ=8tan⁡(42∘)YZ = \frac{8}{\tan(42^{\circ})} YZ≈80.9004YZ \approx \frac{8}{0.9004} YZ≈8.88 cmYZ \approx 8.88 \text{ cm}

Explanation:

Relative to the 42∘42^{\circ} angle at ZZ, XYXY is the opposite side and YZYZ is the adjacent side. We use the tangent ratio (T=O/AT = O/A) and rearrange the formula to solve for the denominator.