A rectangular field with area 63 m² is to be fenced on three sides, with the fourth side being an existing wall. If 25 m of fencing are used, what are the dimensions of the field?

Let the length of the rectangular field be $L$ and the width be $W$.
The area of the field is $A = L \times W = 63 \text{ m}^2$.
The fencing is for three sides: one length and two widths, as one longer side is an existing wall. So, $L > W$.
The total fencing used is $L + 2W = 25 \text{ m}$.

From the area equation, $L = \frac{63}{W}$.
Substitute this into the fencing equation:

Multiply by $W$ to clear the denominator:

Rearrange into a quadratic equation:

Using the quadratic formula $W = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

Two possible values for $W$:

If $W = 9 \text{ m}$, then $L = \frac{63}{9} = 7 \text{ m}$. In this case, $L < W$, which contradicts the condition that the wall is one of the longer sides.
If $W = 3.5 \text{ m}$, then $L = \frac{63}{3.5} = 18 \text{ m}$. In this case, $L > W$, which satisfies the condition.
The dimensions of the field are $18 \text{ m}$ by $3.5 \text{ m}$.

The dimensions of the field are $\boxed{18 \text{ m by } 3.5 \text{ m}}$.

10. (b)

Let $H$ be the height of the building.
Let $D$ be the distance from point $P$ to the foot of the building, $D = 75 \text{ m}$.
The angle of depression from the top of the building to point $P$ is $33.6^\circ$. This means the angle of elevation from point $P$ to the top of the building is also $33.6^\circ$.
We have a right-angled triangle where $H$ is the opposite side to the angle and $D$ is the adjacent side.