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Trigonometry - Sine and Cosine Rules

Grade 12IGCSE

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

🔑Concepts

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The Sine Rule is used to find missing sides or angles when we know either two angles and one side (AAS/ASA) or two sides and a non-included angle (SSA).

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The Cosine Rule is used when we have two sides and the included angle (SAS) to find the third side, or when we have all three sides (SSS) to find an angle.

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The Ambiguous Case of the Sine Rule: When using the Sine Rule to find an angle (SSA), there may be two possible triangles (one acute and one obtuse) if the given side opposite the angle is shorter than the other given side but longer than the altitude.

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Area of a non-right-angled triangle can be calculated using two sides and the sine of the included angle.

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In 3D trigonometry problems, identify right-angled triangles within the 3D shape and use the Sine/Cosine rules to bridge measurements between different planes.

📐Formulae

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

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

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

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

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

💡Examples

Problem 1:

In triangle ABC, side a=12a = 12 cm, side c=15c = 15 cm, and angle B=70∘B = 70^\circ. Find the length of side bb and the area of the triangle.

Solution:

b2=122+152−2(12)(15)cos⁡70∘  ⟹  b2=144+225−360(0.342)  ⟹  b2=245.88  ⟹  b≈15.68b^2 = 12^2 + 15^2 - 2(12)(15) \cos 70^\circ \implies b^2 = 144 + 225 - 360(0.342) \implies b^2 = 245.88 \implies b \approx 15.68 cm. Area =12(12)(15)sin⁡70∘=90(0.9397)≈84.57= \frac{1}{2}(12)(15) \sin 70^\circ = 90(0.9397) \approx 84.57 cm².

Explanation:

Since we are given two sides and the included angle (SAS), we use the Cosine Rule to find the missing side. The area is found using the formula 12acsin⁡B\frac{1}{2}ac \sin B.

Problem 2:

In triangle PQR, PQ=8PQ = 8 cm, QR=10QR = 10 cm, and angle P=60∘P = 60^\circ. Find angle RR.

Solution:

sin⁡R8=sin⁡60∘10  ⟹  sin⁡R=8×sin⁡60∘10  ⟹  sin⁡R=8×0.86610=0.6928  ⟹  R=arcsin⁡(0.6928)≈43.9∘\frac{\sin R}{8} = \frac{\sin 60^\circ}{10} \implies \sin R = \frac{8 \times \sin 60^\circ}{10} \implies \sin R = \frac{8 \times 0.866}{10} = 0.6928 \implies R = \arcsin(0.6928) \approx 43.9^\circ.

Explanation:

We use the Sine Rule because we have a known angle-side pair (PP and QRQR) and want to find an angle opposite a known side (PQPQ).

Problem 3:

Find the largest angle in a triangle with side lengths 5 cm, 7 cm, and 10 cm.

Solution:

Let a=10,b=5,c=7a=10, b=5, c=7. cos⁡A=52+72−1022(5)(7)=25+49−10070=−2670=−0.3714\cos A = \frac{5^2 + 7^2 - 10^2}{2(5)(7)} = \frac{25 + 49 - 100}{70} = \frac{-26}{70} = -0.3714. A=arccos⁡(−0.3714)≈111.8∘A = \arccos(-0.3714) \approx 111.8^\circ.

Explanation:

The largest angle is always opposite the longest side. We use the rearranged Cosine Rule (SSS) to find the angle opposite the 10 cm side.