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Trigonometry - Sine and Cosine Rules

Grade 10IGCSE

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

🔑Concepts

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Standard notation for triangles: Vertices are capital letters (A, B, C) and sides opposite them are lowercase letters (a, b, c).

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The Sine Rule is used when we know a side and its opposite angle, plus one other piece of information (Angle-Angle-Side or Side-Side-Angle).

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The Cosine Rule is used when we have 'Side-Angle-Side' (SAS) to find the third side, or 'Side-Side-Side' (SSS) to find an angle.

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The area of a non-right-angled triangle can be found using any two sides and the included angle (the angle between them).

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For the Sine Rule, be aware of the 'Ambiguous Case' where two possible triangles can exist (though often simplified in IGCSE Core).

📐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, side b=12b = 12 cm, angle B=40∘B = 40^\circ, and angle C=75∘C = 75^\circ. Calculate the length of side cc.

Solution:

  1. Use the Sine Rule: csin⁡75∘=12sin⁡40∘\frac{c}{\sin 75^\circ} = \frac{12}{\sin 40^\circ}.
  2. Rearrange: c=12×sin⁡75∘sin⁡40∘c = \frac{12 \times \sin 75^\circ}{\sin 40^\circ}.
  3. Calculate: c≈18.03c \approx 18.03 cm.

Explanation:

We use the Sine Rule because we have a 'known pair' (side bb and angle BB) and we need to find a side corresponding to another known angle (CC).

Problem 2:

In triangle PQR, pq=7pq = 7 cm, pr=10pr = 10 cm, and angle P=62∘P = 62^\circ. Find the length of side qrqr.

Solution:

  1. Use formula: p2=72+102−2(7)(10)cos⁡62∘p^2 = 7^2 + 10^2 - 2(7)(10) \cos 62^\circ.
  2. p2=49+100−140(0.4695)p^2 = 49 + 100 - 140(0.4695).
  3. p2=149−65.73=83.27p^2 = 149 - 65.73 = 83.27.
  4. p=83.27≈9.13p = \sqrt{83.27} \approx 9.13 cm.

Explanation:

We use the Cosine Rule because we have two sides and the included angle (SAS). Here, let p=qrp = qr, q=10q = 10, and r=7r = 7.

Problem 3:

A triangle has sides of length 5 cm, 8 cm, and 11 cm. Calculate the size of the largest angle.

Solution:

  1. Let a=11a = 11, b=5b = 5, and c=8c = 8.
  2. cos⁡A=52+82−1122(5)(8)\cos A = \frac{5^2 + 8^2 - 11^2}{2(5)(8)}.
  3. cos⁡A=25+64−12180=−3280=−0.4\cos A = \frac{25 + 64 - 121}{80} = \frac{-32}{80} = -0.4.
  4. A=cos⁡−1(−0.4)≈113.6∘A = \cos^{-1}(-0.4) \approx 113.6^\circ.

Explanation:

The largest angle is always opposite the longest side. We use the Cosine Rule for angles (SSS).