IP EMaths Notes (Upper Sec, Year 3-4): 10) Trigonometric Applications

Extend beyond right triangles with the sine rule, cosine rule, and \(\tfrac{1}{2} ab \sin C\). Always sketch the triangle or bearing diagram first to avoid angle ambiguity.

Formula summary

• Sine rule: \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\).
• Cosine rule: \(c^{2} = a^{2} + b^{2} - 2ab \cos C\).
• Area: \(\Delta = \tfrac{1}{2} ab \sin C\). In triangle \(ABC\), sides \(\pu{AB = 7.5 cm}\) and \(\pu{AC = 6.2 cm}\) with included angle \(\pu{BAC = 58 ^\circ}\). Find the length of BC.

1. Apply the cosine rule, working in centimetres:
   \[ BC^2 = (7.5)^2 + (6.2)^2 - 2(7.5)(6.2)\cos\bigl(58^\circ\bigr). \]
2. Square the sides: \(7.5^2 = 56.25\) and \(6.2^2 = 38.44\) (both in \(\text{cm}^2\)).
3. Evaluate the product term:
   \[ 2(7.5)(6.2)\cos\bigl(58^\circ\bigr) \approx 93.0 \times 0.5299 \approx 49.28. \]
4. Substitute the values back into the equation:
   \[ BC^2 \approx 56.25 + 38.44 - 49.28 = 45.41. \]
5. Take the square root to find the side length:
   \[ BC = \sqrt{45.41} \approx 6.74. \]
   Therefore \( BC \approx \pu{6.74 cm} \).

Two ships leave a harbour at the same time. Ship P sails 45 km on a bearing of 040 degrees while ship Q sails 52 km on a bearing of 120 degrees. Estimate their distance apart after the journey.