Mr Ali, Ms Brown and a group of students were camping at the site located at P - HSC - SSCE Mathematics Advanced - Question 15 - 2020 - Paper 1

Using the cosine rule, we can find the distance AB in triangle APB:

Given:

• AP = 7 km
• PB = 9 km
• angle APB = 65°

The cosine rule states: $AB^2 = AP^2 + PB^2 - 2 \times AP \times PB \times \cos(APB)$

Substituting the values:

$AB^2 = 7^2 + 9^2 - 2 \times 7 \times 9 \times \cos(65°)$

Calculating:

• $AB^2 = 49 + 81 - 2 \times 7 \times 9 \times 0.4226$ (cosine value of 65°)
• $AB^2 = 130 - 53.025\approx 76.975$

Taking the square root: $AB \approx \sqrt{76.975} \approx 8.76 \text{ km}$

Thus, the distance AB is approximately 8.76 km.

First, we need to find angle A in triangle APB:

Using the sine rule:

$\frac{sin A}{AB} = \frac{sin 65°}{PB}$

Rearranging gives us: $sin A = \frac{AB \times sin 65°}{PB}$ Substituting the values:

• $AB \approx 8.76 km$
• $PB = 9 km$

Thus: $sin A = \frac{8.76 \times sin 65°}{9} \approx \frac{8.76 \times 0.9063}{9}\approx 0.8768$

Calculating angle A: $A \approx 68.6°$ (using inverse sine)

Finally, the bearing of Ms Brown's group from Mr Ali's group is calculated as: $Bearing = 180° - (A + 35°) \approx 180° - (68.6° + 35°) \approx 180° - 103.6° = 76.4° \approx 146°$

Thus, the bearing of Ms Brown's group from Mr Ali's group is approximately 146°.