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Trigonometry - Sine and Cosine rules

Grade 11IGCSE

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

🔑Concepts

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Labeling triangles: Angles are denoted by capital letters (A, B, C) and sides opposite to them by lowercase letters (a, b, c).

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Non-right angled trigonometry: These rules apply to any triangle, not just right-angled ones.

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The Sine Rule: Used when we have a 'matching pair' (a side and its opposite angle) and one other piece of information.

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The Cosine Rule: Used when we have two sides and the included angle (SAS) or all three sides (SSS).

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The Ambiguous Case: When using the Sine Rule to find an angle (SSA), there may be two possible solutions (acute and obtuse) because sin⁡θ=sin⁡(180∘−θ)\sin \theta = \sin(180^\circ - \theta).

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Area Formula: Calculating the area of a triangle without knowing the vertical height.

📐Formulae

asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} (Sine Rule for finding sides)

sin⁡Aa=sin⁡Bb=sin⁡Cc\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} (Sine Rule for finding angles)

a2=b2+c2−2bccos⁡Aa^2 = b^2 + c^2 - 2bc \cos A (Cosine Rule for finding sides)

cos⁡A=b2+c2−a22bc\cos A = \frac{b^2 + c^2 - a^2}{2bc} (Cosine Rule for finding angles)

Area=12absin⁡C\text{Area} = \frac{1}{2} ab \sin C (Area of any triangle)

💡Examples

Problem 1:

In triangle ABC, side a=12a = 12 cm, angle A=40∘A = 40^\circ, and angle B=60∘B = 60^\circ. Calculate the length of side bb.

Solution:

b=12×sin⁡60∘sin⁡40∘≈16.17b = \frac{12 \times \sin 60^\circ}{\sin 40^\circ} \approx 16.17 cm

Explanation:

Since we have a side-angle pair (aa and AA) and want to find side bb given angle BB, we use the Sine Rule: bsin⁡B=asin⁡A\frac{b}{\sin B} = \frac{a}{\sin A}.

Problem 2:

In triangle PQR, PQ=7PQ = 7 cm, QR=10QR = 10 cm, and PR=8PR = 8 cm. Find the size of angle QQ.

Solution:

cos⁡Q=72+102−822×7×10=85140\cos Q = \frac{7^2 + 10^2 - 8^2}{2 \times 7 \times 10} = \frac{85}{140}. Q=cos⁡−1(0.6071)≈52.6∘Q = \cos^{-1}(0.6071) \approx 52.6^\circ

Explanation:

When three sides are given (SSS), use the Cosine Rule rearranged for the angle. Here, side qq is PR=8PR = 8 cm, and the adjacent sides are p=10p=10 and r=7r=7.

Problem 3:

Find the area of a triangle where two sides are 5 cm and 9 cm, and the included angle is 35∘35^\circ.

Solution:

Area=12×5×9×sin⁡35∘≈12.91\text{Area} = \frac{1}{2} \times 5 \times 9 \times \sin 35^\circ \approx 12.91 cm²

Explanation:

Use the Area formula 12absin⁡C\frac{1}{2}ab \sin C where aa and bb are the given sides and CC is the angle between them.