Solve triangle ABC given angles B = 21°, C = 46°, and side c = 9 cm. Also, determine the true bearing for Diagram 1.

The sum of angles in a triangle is $180^\circ$.
Given $B = 21^\circ$ and $C = 46^\circ$.

The Sine Rule states $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.
We are given $c = AB = 9$ cm.

Using approximate values: $\sin 113^\circ \approx 0.9205$ and $\sin 46^\circ \approx 0.7193$.

Using approximate values: $\sin 21^\circ \approx 0.3584$ and $\sin 46^\circ \approx 0.7193$.

The solved triangle has:
Angle $A = \boxed{113^\circ}$
Side $a \approx \boxed{11.52 \text{ cm}}$
Side $b \approx \boxed{4.48 \text{ cm}}$

Q9: True Bearing of the following diagrams

The angle is $60^\circ$ clockwise from North.
True Bearing = $\boxed{060^\circ}$

The angle is $20^\circ$ from South towards West. True bearing is measured clockwise from North.
From North to South is $180^\circ$. Adding $20^\circ$ towards West from South gives:
True Bearing = $180^\circ + 20^\circ = \boxed{200^\circ}$

The angle is $30^\circ$ from North towards West (counter-clockwise from North).
To find the true bearing (clockwise from North):
True Bearing = $360^\circ - 30^\circ = \boxed{330^\circ}$

Q10: Man's displacement

The man travels West for 80 Km from town A.
His position is 80 Km West of A.

He then travels East for 100 Km. This movement is in the opposite direction to the first.

Let West be negative and East be positive.
Net displacement = $-80 \text{ Km} + 100 \text{ Km}$
Net displacement = $20 \text{ Km}$

Since the net displacement is positive, he is East of town A.
He is $\boxed{20 \text{ Km East}}$ from town A.