Use of Sine, Cosine, and Tangent Ratios

The sine, cosine, and tangent ratios are essential tools for solving problems involving right-angled triangles. These ratios, along with the sine and cosine rules, allow for the calculation of unknown sides and angles in triangles.

Key Concepts

• Sine, Cosine, and Tangent Ratios:
  • \( \sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} \), \( \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} \), \( \tan\theta = \frac{\text{opposite}}{\text{adjacent}} \).
  • Used to find sides and angles of right-angled triangles.
• Sine Rule:
  • \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \).
  • Applies to any triangle, not limited to right-angled triangles.
• Cosine Rule:
  • \( c^2 = a^2 + b^2 – 2ab\cos C \).
  • To find angles: \( \cos C = \frac{a^2 + b^2 – c^2}{2ab} \).
• Area of a Triangle:
  • \( \text{Area} = \frac{1}{2}ab\sin C \).
  • Works for any triangle where two sides and the included angle are known.

Guidance, Clarifications, and Syllabus Links

• Students should always sketch well-labelled diagrams to visualize and support solutions.
• Relates to inverse functions (SL 2.2) when finding angles using trigonometric ratios.
• This section excludes the ambiguous case of the sine rule.

Examples

Example 1: Using Trigonometric Ratios

• Find the length of side \( x \) in a right-angled triangle where \( \theta = 30^\circ \) and the hypotenuse is 10 cm:
  • \( \sin 30^\circ = \frac{x}{10} \).
  • \( x = 10 \cdot \sin 30^\circ = 10 \cdot 0.5 = 5 \, \text{cm} \).

Example 2: Applying the Sine Rule

• Find angle \( B \) in a triangle where \( a = 8 \), \( b = 10 \), and \( A = 45^\circ \):
  • \( \frac{a}{\sin A} = \frac{b}{\sin B} \).
  • \( \frac{8}{\sin 45^\circ} = \frac{10}{\sin B} \).
  • \( \sin B = \frac{10 \cdot \sin 45^\circ}{8} = \frac{10 \cdot 0.707}{8} = 0.884 \).
  • \( B = \arcsin(0.884) = 62.2^\circ \).

Example 3: Using the Cosine Rule

• Find side \( c \) in a triangle where \( a = 7 \), \( b = 9 \), and \( C = 60^\circ \):
  • \( c^2 = a^2 + b^2 – 2ab\cos C \).
  • \( c^2 = 7^2 + 9^2 – 2 \cdot 7 \cdot 9 \cdot \cos 60^\circ \).
  • \( c^2 = 49 + 81 – 63 = 67 \).
  • \( c = \sqrt{67} \approx 8.19 \, \text{units} \).

Example 4: Area of a Triangle

• Find the area of a triangle where \( a = 6 \), \( b = 8 \), and \( C = 90^\circ \):
  • \( \text{Area} = \frac{1}{2}ab\sin C \).
  • \( \text{Area} = \frac{1}{2} \cdot 6 \cdot 8 \cdot \sin 90^\circ = 24 \, \text{square units} \).