John starts at O. He travels 40 km on a bearing of 210° to town T. Attah then travels 30 km from T on a bearing of 300° to his final position A. Find the coordinates of A relative to O.

a)

Step 1: Determine the coordinates of town T from John's initial position.
Let John's initial position be the origin $O(0,0)$. John travels 40 km on a bearing of $210^\circ$.
To convert a bearing to a standard angle $\theta$ (measured counter-clockwise from the positive x-axis), we use $\theta = 90^\circ - \text{bearing}$.
For a bearing of $210^\circ$:

$$\theta_T = 90^\circ - 210^\circ = -120^\circ$$

The coordinates of T are $(x_T, y_T)$:

$$x_T = 40 \cos(-120^\circ) = 40 \left(-\frac{1}{2}\right) = -20$$ $$y_T = 40 \sin(-120^\circ) = 40 \left(-\frac{\sqrt{3}}{2}\right) = -20\sqrt{3}$$

So, the coordinates of town T are $(-20, -20\sqrt{3})$.

Step 2: Determine the displacement of Attah from town T.
Attah moves 30 km from T on a bearing of $300^\circ$.
For a bearing of $300^\circ$:

$$\theta_A = 90^\circ - 300^\circ = -210^\circ$$

The displacement vector from T to Attah's final position is $(x_{disp}, y_{disp})$:

$$x_{disp} = 30 \cos(-210^\circ) = 30 \cos(150^\circ) = 30 \left(-\frac{\sqrt{3}}{2}\right) = -15\sqrt{3}$$ $$y_{disp} = 30 \sin(-210^\circ) = 30 \sin(150^\circ) = 30 \left(\frac{1}{2}\right) = 15$$

Step 3: Calculate Attah's final position from John's initial position.
Attah's final position $(x_{Attah}, y_{Attah})$ is the sum of T's coordinates and Attah's displacement from T:

$$x_{Attah} = x_T + x_{disp} = -20 - 15\sqrt{3}$$ $$y_{Attah} = y_T + y_{disp} = -20\sqrt{3} + 15$$

Using $\sqrt{3} \approx 1.732$:

$$x_{Attah} = -20 - 15(1.732) = -20 - 25.98 = -45.98$$ $$y_{Attah} = -20(1.732) + 15 = -34.64 + 15 = -19.64$$

The position of Attah from John's initial position is approximately $(-45.98, -19.64)$ km.

The position of Attah from John's initial position is $\boxed{(-45.98, -19.64)}$.

b)
Given $A(1,2)$ and the vectors $\vec{AB} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$, $\vec{BC} = \begin{pmatrix} -1 \\ 3 \end{pmatrix}$, and $\vec{CD} = \begin{pmatrix} 2 \\ -2 \end{pmatrix}$.

Step 1: Calculate the coordinates of B.
The position vector of B is $\vec{OB} = \vec{OA} + \vec{AB}$.

$$\vec{OB} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + \begin{pmatrix} 3 \\ 2 \end{pmatrix} = \begin{pmatrix} 1+3 \\ 2+2 \end{pmatrix} = \begin{pmatrix} 4 \\ 4 \end{pmatrix}$$

So, the coordinates of B are $(4,4)$.

Step 2: Calculate the coordinates of C.
The position vector of C is $\vec{OC} = \vec{OB} + \vec{BC}$.

$$\vec{OC} = \begin{pmatrix} 4 \\ 4 \end{pmatrix} + \begin{pmatrix} -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 4-1 \\ 4+3 \end{pmatrix} = \begin{pmatrix} 3 \\ 7 \end{pmatrix}$$

So, the coordinates of C are $(3,7)$.

Step 3: Calculate the coordinates of D.
The position vector of D is $\vec{OD} = \vec{OC} + \vec{CD}$.

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