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Geometry and Trigonometry - The Sine Rule, Cosine Rule, and Area of a Triangle

Grade 11IB

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

🔑Concepts

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Standard Triangle Labeling: In any non-right triangle ABCABC, the vertices and their corresponding interior angles are denoted by capital letters A,B,A, B, and CC. The side lengths opposite these angles are denoted by lowercase letters a,b,a, b, and cc. Visually, side aa is the segment that connects vertices BB and CC, directly across from angle AA.

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The Sine Rule: This rule establishes a proportional relationship between the side lengths of a triangle and the sines of their opposite angles. It is most effectively used when you know an 'angle-side pair' (an angle and the side opposite it) plus one other piece of information. Visually, it links the ratio of any side to its 'facing' angle.

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The Ambiguous Case of the Sine Rule: When you are given two sides and a non-included acute angle (Side-Side-Angle or SSASSA), there is a possibility of two different triangles existing. Visually, imagine one of the given sides acting as a pivot; if it is shorter than the other given side but longer than the triangle's height, it can intersect the base at two distinct points, creating one triangle with an acute angle and another with an obtuse angle.

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The Cosine Rule for Sides: This rule is used to find a missing side when you know two sides and the included angle (the angle between them, SASSAS). Visually, it functions like a generalized version of the Pythagorean theorem, where the term −2bccos⁡A-2bc \cos A adjusts for the fact that the triangle is not right-angled.

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The Cosine Rule for Angles: By rearranging the side formula, the Cosine Rule can find any interior angle if all three side lengths are known (SSSSSS). Visually, this means the 'shape' of the triangle is fixed by its side lengths, and the formula extracts the specific opening of the angle between any two sides.

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Area of a Triangle using Sine: The area of any triangle can be calculated using two side lengths and the sine of the angle trapped between them (SASSAS). Visually, this is equivalent to 12×base×height\frac{1}{2} \times \text{base} \times \text{height}, where the height is represented by bsin⁡Ab \sin A.

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Solving Non-Right Triangles: To solve a triangle (find all missing parts), choose the rule based on the given information. Use the Sine Rule for AASAAS or SSASSA scenarios, and use the Cosine Rule for SASSAS or SSSSSS scenarios. Visually, always sketch the triangle first to identify which sides are opposite which angles.

📐Formulae

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

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

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

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

Area=12absin⁡C\text{Area} = \frac{1}{2}ab \sin C (Area formula using sides a,ba, b and angle CC)

💡Examples

Problem 1:

In triangle ABCABC, angle A=40∘A = 40^{\circ}, angle B=60∘B = 60^{\circ}, and side a=12a = 12 cm. Find the length of side bb to two decimal places.

Solution:

Step 1: Identify the given information. We have an angle-side pair (AA and aa) and another angle (BB). This requires the Sine Rule. Step 2: Set up the ratio: bsin⁡B=asin⁡A\frac{b}{\sin B} = \frac{a}{\sin A}. Step 3: Substitute the values: bsin⁡60∘=12sin⁡40∘\frac{b}{\sin 60^{\circ}} = \frac{12}{\sin 40^{\circ}}. Step 4: Isolate bb: b=12×sin⁡60∘sin⁡40∘b = \frac{12 \times \sin 60^{\circ}}{\sin 40^{\circ}}. Step 5: Calculate the result: b≈12×0.86600.6428≈16.17b \approx \frac{12 \times 0.8660}{0.6428} \approx 16.17 cm.

Explanation:

Since we were given two angles and one opposite side, the Sine Rule is the most direct method to find the missing side length.

Problem 2:

In triangle PQRPQR, side p=7p = 7 cm, side q=10q = 10 cm, and side r=15r = 15 cm. Find the measure of angle RR to the nearest degree.

Solution:

Step 1: Identify the given information. We have three sides (SSSSSS), so we must use the Cosine Rule rearranged for an angle. Step 2: Use the formula cos⁡R=p2+q2−r22pq\cos R = \frac{p^2 + q^2 - r^2}{2pq}. Step 3: Substitute the values: cos⁡R=72+102−1522(7)(10)\cos R = \frac{7^2 + 10^2 - 15^2}{2(7)(10)}. Step 4: Simplify the expression: cos⁡R=49+100−225140=−76140≈−0.5429\cos R = \frac{49 + 100 - 225}{140} = \frac{-76}{140} \approx -0.5429. Step 5: Find the inverse cosine: R=cos⁡−1(−0.5429)≈122.88∘R = \cos^{-1}(-0.5429) \approx 122.88^{\circ}. Step 6: Round to the nearest degree: R≈123∘R \approx 123^{\circ}.

Explanation:

When all three sides are known, the Cosine Rule is the only way to determine the angles. A negative value for cos⁡R\cos R correctly indicates that angle RR is obtuse.