Question:

1. Draw a diagram (not shown). Let:
- \(P\) be the top of the multi‑storeyed building (height \(H\) from ground).
- \(T\) be the top of the 8 m building.
- \(B\) be the bottom (ground) of the 8 m building.
- \(d\) be the horizontal distance between the two buildings.

2. Convert angles of depression to angles of elevation:
- Angle of depression to the top \(T\) = 30° ⇒ angle of elevation from \(T\) to \(P\) = 30°.
- Angle of depression to the bottom \(B\) = 45° ⇒ angle of elevation from \(B\) to \(P\) = 45°.

3. Apply the tangent definition to the two right‑angled triangles:
- For triangle \(P T\) (top to top):
$$\tan 30^{\circ}=\frac{\text{vertical difference}}{\text{horizontal distance}}=\frac{H-8}{d}$$
Since \(\tan30^{\circ}=\frac{1}{\sqrt3}\),
$$\frac{H-8}{d}=\frac{1}{\sqrt3}\quad\Rightarrow\quad H-8=\frac{d}{\sqrt3}\tag{1}$$
- For triangle \(P B\) (top to bottom):
$$\tan 45^{\circ}=\frac{H}{d}$$
Because \(\tan45^{\circ}=1\),
$$\frac{H}{d}=1\quad\Rightarrow\quad H=d\tag{2}$$

4. Substitute \(H=d\) from (2) into (1):
$$d-8=\frac{d}{\sqrt3}\
\Rightarrow\; \sqrt3\,(d-8)=d\
\Rightarrow\; \sqrt3\,d-8\sqrt3=d\
\Rightarrow\; d(\sqrt3-1)=8\sqrt3$$

5. Solve for \(d\):
$$d=\frac{8\sqrt3}{\sqrt3-1}\
=\frac{8\sqrt3(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}\
=\frac{8\sqrt3(\sqrt3+1)}{3-1}\
=4\sqrt3(\sqrt3+1)\
=4(3+\sqrt3)\
=12+4\sqrt3\;\text{metres}$$
Numerically, \(d\approx12+4(1.732)=12+6.928\approx18.93\) m.

6. Find the height \(H\) using (2):
$$H=d=12+4\sqrt3\;\text{metres}\approx18.93\;\text{m}$$

7. Answer: Height of the multi‑storeyed building = \(12+4\sqrt3\) m (≈ 18.93 m). Distance between the two buildings = \(12+4\sqrt3\) m (≈ 18.93 m).