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Trigonometry - Sine and Cosine Rules

Grade 9IGCSE

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

🔑Concepts

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Standard notation: In a triangle ABC, side 'a' is opposite angle A, side 'b' is opposite angle B, and side 'c' is opposite angle C.

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The Sine Rule is used when you have a matching pair of an angle and its opposite side (e.g., Angle A and side a).

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The Sine Rule can find a missing side if two angles and one side are known, or a missing angle if two sides and one non-included angle are known.

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The Cosine Rule is used for SAS (Side-Angle-Side) to find the third side.

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The Cosine Rule is used for SSS (Side-Side-Side) to find any missing angle.

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The Area of a Triangle formula is used when you know two sides and the 'included' angle between them.

📐Formulae

Sine Rule (to find a side): asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

Sine Rule (to find an angle): sin⁡Aa=sin⁡Bb=sin⁡Cc\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}

Cosine Rule (to find a side): a2=b2+c2−2bccos⁡Aa^2 = b^2 + c^2 - 2bc \cos A

Cosine Rule (to find an angle): cos⁡A=b2+c2−a22bc\cos A = \frac{b^2 + c^2 - a^2}{2bc}

Area of a Triangle: Area=12absin⁡C\text{Area} = \frac{1}{2}ab \sin C

💡Examples

Problem 1:

In triangle ABC, angle A = 40°, angle B = 60°, and side a = 12 cm. Calculate the length of side b.

Solution:

  1. Use the Sine Rule: bsin⁡60∘=12sin⁡40∘\frac{b}{\sin 60^\circ} = \frac{12}{\sin 40^\circ}
  2. Rearrange: b=12×sin⁡60∘sin⁡40∘b = \frac{12 \times \sin 60^\circ}{\sin 40^\circ}
  3. Calculate: b=12×0.86600.6428≈16.17b = \frac{12 \times 0.8660}{0.6428} \approx 16.17 cm.

Explanation:

We use the Sine Rule because we have a known angle-side pair (A and a) and we are looking for a side opposite a known angle (B).

Problem 2:

In triangle PQR, PQ = 7 cm, QR = 10 cm, and the angle PQR = 75°. Find the length of side PR.

Solution:

  1. Let p=10p=10, r=7r=7, and angle Q=75∘Q=75^\circ. Use Cosine Rule: q2=p2+r2−2prcos⁡Qq^2 = p^2 + r^2 - 2pr \cos Q
  2. Substitute: q2=102+72−2(10)(7)cos⁡75∘q^2 = 10^2 + 7^2 - 2(10)(7) \cos 75^\circ
  3. q2=100+49−140(0.2588)=149−36.23=112.77q^2 = 100 + 49 - 140(0.2588) = 149 - 36.23 = 112.77
  4. q=112.77≈10.62q = \sqrt{112.77} \approx 10.62 cm.

Explanation:

We use the Cosine Rule because we have two sides and the 'included' angle (SAS). This configuration does not provide a complete angle-side pair for the Sine Rule.

Problem 3:

A triangle has sides of length 5 cm, 8 cm, and 9 cm. Calculate the size of the smallest angle.

Solution:

  1. The smallest angle is opposite the shortest side (5 cm). Let a=5,b=8,c=9a=5, b=8, c=9.
  2. Use Cosine Rule for angle A: cos⁡A=82+92−522(8)(9)\cos A = \frac{8^2 + 9^2 - 5^2}{2(8)(9)}
  3. cos⁡A=64+81−25144=120144=0.8333\cos A = \frac{64 + 81 - 25}{144} = \frac{120}{144} = 0.8333
  4. A=cos⁡−1(0.8333)≈33.6∘A = \cos^{-1}(0.8333) \approx 33.6^\circ.

Explanation:

When three sides are given (SSS), the Cosine Rule is required to find any interior angle.