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Geometry and Trigonometry - Applications of Trigonometry (Sine and Cosine Rules)

Grade 12IB

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

🔑Concepts

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Standard Triangle Labeling: In a non-right-angled triangle ABCABC, angles are represented by capital letters and the sides opposite them by corresponding lower-case letters aa, bb, and cc. Visually, side aa is across from vertex AA, side bb across from BB, and side cc across from CC. Identifying these pairs is the first step in applying any trigonometric rule.

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The Sine Rule: This rule states that the ratio of a side's length to the sine of its opposite angle is constant for all sides in a triangle. It is primarily used when you know a side and its opposite angle, plus one other side or angle (AAS or SSA cases). Visually, as an angle increases, its opposite side length also increases in proportion to its sine.

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The Ambiguous Case (Sine Rule): When given two sides and a non-included angle (SSA), there may be zero, one, or two possible triangles. This occurs visually when the side opposite the given angle is shorter than the other given side but longer than the triangle's altitude (h=bsinAh = b \\sin A); the side can 'swing' like a pendulum to touch the base at two different points, creating one acute and one obtuse triangle.

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The Cosine Rule (Finding Sides): This is used when you are given two sides and the included angle (SAS). It functions as a generalized version of the Pythagorean theorem. Visually, imagine the two sides as the arms of a hinge; the Cosine Rule calculates the gap between the ends of the arms based on the angle of the hinge.

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The Cosine Rule (Finding Angles): When all three sides of a triangle are known (SSS), the Cosine Rule can be rearranged to find any angle. In a triangle with sides a,b,ca, b, c, the largest angle will always be opposite the longest side, which is a helpful visual check during calculations.

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Area of a Non-Right Triangle: The area is calculated using two sides and the sine of the included angle. Visually, this formula represents half the area of a parallelogram formed by the two known sides; the 'opening' (angle) between the sides determines the height and thus the total area.

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Bearings and Navigation: Bearings are angles measured clockwise from North (000circ000^{\\circ}). In trigonometry applications, visual diagrams are essential; drawing a North line at each vertex allows the use of parallel line theorems (like alternate interior angles) to calculate the internal angles of the triangle formed by the travel path.

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3D Trigonometry Applications: These problems involve finding lengths and angles in 3D solids like pyramids or cuboids. The strategy is to identify 2D 'slices' within the 3D object to form right-angled or non-right-angled triangles. Visually, look for diagonals that span across the interior of the shape to bridge different faces.

📐Formulae

Sine Rule: fracasinA=fracbsinB=fraccsinC\\frac{a}{\\sin A} = \\frac{b}{\\sin B} = \\frac{c}{\\sin C}

Sine Rule (Rearranged for Angles): fracsinAa=fracsinBb=fracsinCc\\frac{\\sin A}{a} = \\frac{\\sin B}{b} = \\frac{\\sin C}{c}

Cosine Rule (Finding a Side): a2=b2+c2−2bccosAa^2 = b^2 + c^2 - 2bc \\cos A

Cosine Rule (Finding an Angle): \cos A = \frac{b^2 + c^2 - a^2}{2bc}$

Area of a Triangle: Area=frac12absinCArea = \\frac{1}{2}ab \\sin C

💡Examples

Problem 1:

In triangle ABCABC, side a=8a = 8 cm, side b=11b = 11 cm, and the included angle C=35circC = 35^{\\circ}. Calculate the length of side cc and the total area of the triangle.

Solution:

  1. Find side cc: Use the Cosine Rule for SAS: c2=a2+b2−2abcosCc^2 = a^2 + b^2 - 2ab \\cos C.
  2. Substitute values: c2=82+112−2(8)(11)cos35circc^2 = 8^2 + 11^2 - 2(8)(11) \\cos 35^{\\circ}.
  3. Calculate: c2=64+121−176(0.8192)approx185−144.18=40.82c^2 = 64 + 121 - 176(0.8192) \\approx 185 - 144.18 = 40.82.
  4. Take the square root: c=sqrt40.82approx6.39c = \\sqrt{40.82} \\approx 6.39 cm.
  5. Find Area: Use the formula Area=frac12absinCArea = \\frac{1}{2}ab \\sin C.
  6. Substitute values: Area=frac12(8)(11)sin35circapprox44(0.5736)=25.24Area = \\frac{1}{2}(8)(11) \\sin 35^{\\circ} \\approx 44(0.5736) = 25.24 cm2^2.

Explanation:

Since we are given two sides and the angle between them (SAS), the Cosine Rule is required to find the third side. The area formula also uses the SAS components directly.

Problem 2:

In triangle PQRPQR, side q=10q = 10 cm, side p=6p = 6 cm, and angle P=30circP = 30^{\\circ}. Determine the two possible values for angle QQ.

Solution:

  1. Set up Sine Rule: fracsinQq=fracsinPp\\frac{\\sin Q}{q} = \\frac{\\sin P}{p}.
  2. Rearrange to solve for sinQ\\sin Q: sinQ=fracqsinPp=frac10sin30circ6\\sin Q = \\frac{q \\sin P}{p} = \\frac{10 \\sin 30^{\\circ}}{6}.
  3. Calculate: sinQ=frac10(0.5)6=frac56approx0.8333\\sin Q = \\frac{10(0.5)}{6} = \\frac{5}{6} \\approx 0.8333.
  4. Find the first possible angle (acute): Q1=arcsin(0.8333)approx56.4circQ_1 = \\arcsin(0.8333) \\approx 56.4^{\\circ}.
  5. Find the second possible angle (obtuse): Q2=180circ−56.4circ=123.6circQ_2 = 180^{\\circ} - 56.4^{\\circ} = 123.6^{\\circ}.
  6. Verify both: P+Q2=30circ+123.6circ=153.6circP + Q_2 = 30^{\\circ} + 123.6^{\\circ} = 153.6^{\\circ}, which is less than 180circ180^{\\circ}, so both are valid.

Explanation:

This is the ambiguous case of the Sine Rule (SSA). Because the side opposite the given angle (p=6p=6) is shorter than the other side (q=10q=10) but longer than the height (10sin30circ=510 \\sin 30^{\\circ} = 5), two distinct triangles can be formed.