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Trigonometry - Sine and Cosine rules

Grade 11A Level

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

🔑Concepts

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The Sine Rule connects the lengths of sides to the sines of their opposite angles. Use asin⁡A=bsin⁡B\frac{a}{\sin A} = \frac{b}{\sin B} to find a missing side, and sin⁡Aa=sin⁡Bb\frac{\sin A}{a} = \frac{\sin B}{b} to find a missing angle. This rule is applied when you know a matching 'side-angle' pair and one other piece of information.

Standard triangle labeled with vertices A, B, C and opposite sides a, b, c.
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The Cosine Rule a2=b2+c2−2bccos⁡Aa^2 = b^2 + c^2 - 2bc \cos A is used to find a third side when two sides and the included angle (SAS) are known. It is also used to find an angle when all three sides (SSS) are known, using the rearranged form cos⁡A=b2+c2−a22bc\cos A = \frac{b^2 + c^2 - a^2}{2bc}.

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Area of a non-right-angled triangle can be calculated using 12absin⁡C\frac{1}{2} ab \sin C. This requires two sides and the angle between them (the 'included' angle).

Diagram showing sides a and b with included angle C.
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The 'Ambiguous Case' occurs when using the Sine Rule with two sides and a non-included angle (SSA). There can be two possible triangles if the side opposite the given angle is shorter than the other given side but longer than the altitude.

📐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.

Problem 4:

In triangle XYZ, XY=15XY = 15 cm, XZ=10XZ = 10 cm, and angle X=110∘X = 110^\circ. Calculate the length of YZYZ.

Obtuse triangle XYZ with X=110, XY=15, and XZ=10.

Solution:

  1. Identify the given information: Two sides and the included angle (SAS). Use the Cosine Rule.
  2. Let xx be the length of YZYZ, y=10y = 10 cm, z=15z = 15 cm, and X=110∘X = 110^\circ.
  3. Substitute into the formula: x2=102+152−2(10)(15)cos⁡(110∘)x^2 = 10^2 + 15^2 - 2(10)(15) \cos(110^\circ)
  4. Calculate the values: x2=100+225−300(−0.3420)x^2 = 100 + 225 - 300(-0.3420)
  5. x2=325+102.6=427.6x^2 = 325 + 102.6 = 427.6
  6. x=427.6≈20.68x = \sqrt{427.6} \approx 20.68
  7. The length of YZYZ is 20.720.7 cm (to 3 s.f.).

Explanation:

Since we are given two sides and the angle between them, the Cosine Rule is the most direct method to find the opposite side.

Problem 5:

In triangle ABC, AC=8AC = 8 m, angle A=45∘A = 45^\circ, and angle C=70∘C = 70^\circ. Find the area of the triangle.

Triangle ABC with side AC=8, angle A=45 and angle C=70.

Solution:

  1. Find angle BB: B=180∘−(45∘+70∘)=65∘B = 180^\circ - (45^\circ + 70^\circ) = 65^\circ
  2. Use the Sine Rule to find side aa (BC): asin⁡45∘=8sin⁡65∘\frac{a}{\sin 45^\circ} = \frac{8}{\sin 65^\circ}
  3. a=8×sin⁡45∘sin⁡65∘≈8×0.70710.9063≈6.241a = \frac{8 \times \sin 45^\circ}{\sin 65^\circ} \approx \frac{8 \times 0.7071}{0.9063} \approx 6.241
  4. Use the Area formula with sides b=8b=8, a=6.241a=6.241 and included angle C=70∘C=70^\circ: Area=12×8×6.241×sin⁡70∘\text{Area} = \frac{1}{2} \times 8 \times 6.241 \times \sin 70^\circ
  5. Area=4×6.241×0.9397≈23.45\text{Area} = 4 \times 6.241 \times 0.9397 \approx 23.45
  6. The area is 23.523.5 m2^2 (to 3 s.f.).

Explanation:

To find the area, we need two sides and the included angle. We calculated the third angle first, then used the Sine Rule to find a second side.