Question 2 (5 Marks)
Mr Ali, Ms Brown and a group of students were camping at the site located at $P$. Mr Ali walked with some of the students on a bearing of $035^{\circ}$ for $7 \mathrm{~km}$ to location A. Ms Brown, with the rest of the students, walked on a bearing of $100^{\circ}$ for $9 \mathrm{~km}$ to location $B$.
a. Show that the angle $A P B$ is $65^{\circ}$. (1 mark)
b. Find the distance $A B$. (2 marks)
c. Find the bearing of Ms Brown's group from Mr Ali's group. Give your answer correct to the nearest degree. (2 marks)

Step 1 ：Let's denote the point at which Mr Ali and his group are located as point $A$, the point at which Ms Brown and her group are located as point $B$, and the camping site as point $P$.

Step 2 ：Since Mr Ali walked on a bearing of $035^{\circ}$ and Ms Brown walked on a bearing of $100^{\circ}$, the angle between their paths is $100^{\circ} - 035^{\circ} = 65^{\circ}$.

Step 3 ：So, the angle $APB$ is $65^{\circ}$, which can be written as \(\boxed{65^{\circ}}\).

Step 4 ：For part b, we can use the Law of Cosines to find the distance $AB$. The Law of Cosines states that $c^2 = a^2 + b^2 - 2ab\cos(C)$, where $a$ and $b$ are the lengths of the sides of a triangle, $c$ is the length of the side opposite angle $C$, and $\cos(C)$ is the cosine of angle $C$.

Step 5 ：In this case, $a = 7\mathrm{~km}$ (the distance Mr Ali and his group walked), $b = 9\mathrm{~km}$ (the distance Ms Brown and her group walked), and $C = 65^{\circ}$ (the angle between their paths).

Step 6 ：Substituting these values into the Law of Cosines gives $AB^2 = 7^2 + 9^2 - 2 \cdot 7 \cdot 9 \cdot \cos(65^{\circ})$.

Step 7 ：Solving this equation gives $AB = \sqrt{49 + 81 - 126 \cdot \cos(65^{\circ})} \approx 5.6\mathrm{~km}$.

Step 8 ：So, the distance $AB$ is approximately \(\boxed{5.6\mathrm{~km}}\).

Step 9 ：For part c, we can use the Law of Sines to find the bearing of Ms Brown's group from Mr Ali's group. The Law of Sines states that $\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}$, where $a$, $b$, and $c$ are the lengths of the sides of a triangle, $A$, $B$, and $C$ are the angles opposite those sides, and $\sin(A)$, $\sin(B)$, and $\sin(C)$ are the sines of those angles.

Step 10 ：In this case, $a = 7\mathrm{~km}$, $b = 9\mathrm{~km}$, $c = 5.6\mathrm{~km}$, $A = 35^{\circ}$, $B = ?$, and $C = 65^{\circ}$.

Step 11 ：Substituting these values into the Law of Sines gives $\frac{7}{\sin(35^{\circ})} = \frac{5.6}{\sin(65^{\circ})}$.

Step 12 ：Solving this equation gives $B = \sin^{-1}\left(\frac{5.6 \cdot \sin(35^{\circ})}{7}\right) \approx 55^{\circ}$.

Step 13 ：So, the bearing of Ms Brown's group from Mr Ali's group is approximately \(\boxed{55^{\circ}}\).