Mr Ali, Ms Brown and a group of students were camping at the site located at P - HSC - SSCE Mathematics Standard - Question 31 - 2020 - Paper 1

To find the distance AB, we can use the cosine rule:

$AB^2 = PA^2 + PB^2 - 2 \cdot PA \cdot PB \cdot \cos(65°)$

Here, we know:

• PA = 7 km
• PB = 9 km

Substituting the values:

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

Calculating:

$AB^2 = 49 + 81 - 126 \cdot 0.4226 = 130 - 53.3896 = 76.6104$

Taking the square root, we find:

$AB = \sqrt{76.6104} \approx 8.76 \text{ km}$

We can use the sine rule to find angle A:

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

Substituting the known values:

$\frac{sin A}{8.76} = \frac{sin 65°}{9}$

Calculating angle A:

$A = \arcsin\left( \frac{sin 65° \cdot 8.76}{9} \right) \approx 68.6°$

Now, we find the bearing:

$\text{Bearing} = 180° - (A + 35°) = 180° - (68.6° + 35°) = 180° - 103.6° = 76.4°$

Rounding to the nearest degree gives a bearing of 146°.