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Geometry and Trigonometry - Sine and Cosine rules

Grade 10IB

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

🔑Concepts

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Standard Triangle Notation: In non-right angled trigonometry, a triangle is labeled with vertices AA, BB, and CC in uppercase. The side lengths opposite these vertices are denoted by the corresponding lowercase letters aa, bb, and cc. Visually, side aa is the segment connecting BB and CC, side bb connects AA and CC, and side cc connects AA and BB.

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The Sine Rule: This rule establishes a proportional relationship between the side lengths and the sine of their opposite angles. It is used primarily in two scenarios: when you know two angles and one side (ASA or AAS) or when you know two sides and a non-included angle (SSA).

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Ambiguous Case of the Sine Rule: When using the Sine Rule with the SSA condition, it is possible that zero, one, or two distinct triangles can be formed. Visually, if the side opposite the given angle is shorter than the other given side but longer than the altitude of the triangle, that side can 'swing' to two different positions, creating one acute and one obtuse triangle.

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The Cosine Rule for Sides: This rule is an extension of the Pythagorean theorem for non-right triangles. It is used when you have the 'Side-Angle-Side' (SAS) configuration, where an angle is 'sandwiched' between two known side lengths. Visually, you use the two sides forming the 'V' shape and the angle between them to calculate the distance of the 'opening' side.

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The Cosine Rule for Angles: By rearranging the side formula, the Cosine Rule can solve for any interior angle if all three side lengths (SSS) are known. This is particularly useful because, unlike the Sine Rule, the Cosine Rule provides a unique solution for angles between 0∘0^\circ and 180∘180^\circ, distinguishing clearly between acute and obtuse angles.

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Area of a General Triangle: Beyond the standard 12×base×height\frac{1}{2} \times \text{base} \times \text{height}, the area can be calculated using any two sides and the sine of the included angle. Visually, if you know the lengths of two sides meeting at a vertex, the area is half the product of those sides multiplied by the sine of the angle they form.

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Solving Strategy Selection: To decide which rule to use, look at the given information. Use the Sine Rule if you have a 'known pair' (a side and its opposite angle). Use the Cosine Rule if you have three sides (SSS) or two sides and the included angle (SAS).

📐Formulae

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

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

a2=b2+c2−2bccos⁡(A)a^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 of a triangle)

💡Examples

Problem 1:

In △ABC\triangle ABC, angle A=40∘A = 40^\circ, angle B=60∘B = 60^\circ, and side a=12 cma = 12 \text{ cm}. Calculate the length of side bb to 2 decimal places.

Solution:

  1. Identify the given information: A=40∘,B=60∘,a=12A = 40^\circ, B = 60^\circ, a = 12. We need to find bb.
  2. Choose the Sine Rule because we have a known side-angle pair (aa and AA): asin⁡(A)=bsin⁡(B)\frac{a}{\sin(A)} = \frac{b}{\sin(B)}
  3. Substitute the values: 12sin⁡(40∘)=bsin⁡(60∘)\frac{12}{\sin(40^\circ)} = \frac{b}{\sin(60^\circ)}
  4. Rearrange to solve for bb: b=12×sin⁡(60∘)sin⁡(40∘)b = \frac{12 \times \sin(60^\circ)}{\sin(40^\circ)}
  5. Calculate: b≈12×0.86600.6428≈16.17 cmb \approx \frac{12 \times 0.8660}{0.6428} \approx 16.17 \text{ cm}

Explanation:

Since we were given two angles and a side (AAS), the Sine Rule is the most direct method. We set up a ratio between the side we want to find and its opposite angle, equating it to the ratio of the side we know and its opposite angle.

Problem 2:

In △PQR\triangle PQR, side p=7 cmp = 7 \text{ cm}, side q=10 cmq = 10 \text{ cm}, and side r=13 cmr = 13 \text{ cm}. Find the measure of angle QQ.

Solution:

  1. Identify the given information: p=7,q=10,r=13p=7, q=10, r=13 (SSS). We need to find angle QQ.
  2. Choose the Cosine Rule for angles: cos⁡(Q)=p2+r2−q22pr\cos(Q) = \frac{p^2 + r^2 - q^2}{2pr}
  3. Substitute the side lengths: cos⁡(Q)=72+132−1022(7)(13)\cos(Q) = \frac{7^2 + 13^2 - 10^2}{2(7)(13)}
  4. Simplify the expression: cos⁡(Q)=49+169−100182\cos(Q) = \frac{49 + 169 - 100}{182} cos⁡(Q)=118182≈0.6484\cos(Q) = \frac{118}{182} \approx 0.6484
  5. Find the inverse cosine: Q=cos⁡−1(0.6484)≈49.6∘Q = \cos^{-1}(0.6484) \approx 49.6^\circ

Explanation:

When all three sides of a triangle are known (SSS), the Cosine Rule is the only way to find an interior angle. Note that the side opposite the angle we are finding (q=10q=10) is the one subtracted in the numerator.