Question:

1. Draw a diagram (described):
- Let \(B\) be the top of the building (height = 7 m above ground).
- Let \(F\) be the foot of the tower on the ground.
- Let \(T\) be the top of the tower (height = \(h\) meters above ground).
- The horizontal line through \(B\) meets the ground at a point directly below \(B\); the line \(BF\) makes an angle of depression \(45^{\circ}\) with this horizontal.
- The line of sight \(BT\) makes an angle of elevation \(60^{\circ}\) with the same horizontal.
- Both \(BF\) and \(BT\) share the same horizontal distance \(x = BF = BT_{\text{horizontal}}\).

2. Use the angle of depression (45°) to find the horizontal distance \(x\):
\[
\tan 45^{\circ} = \frac{\text{vertical drop}}{\text{horizontal distance}} = \frac{7}{x}
\]
Since \(\tan 45^{\circ}=1\), we get \(1 = \frac{7}{x}\) ⇒ \(x = 7\) metres.

3. Use the angle of elevation (60°) to relate \(x\) to the height of the tower:
The vertical difference between the tower top and the building top is \(h-7\). Hence,
\[
\tan 60^{\circ} = \frac{h-7}{x}
\]
With \(\tan 60^{\circ}=\sqrt{3}\) and \(x = 7\),
\[
\sqrt{3} = \frac{h-7}{7}
\]
Multiply both sides by 7:
\[
h - 7 = 7\sqrt{3}
\]
Therefore,
\[
h = 7 + 7\sqrt{3} = 7(1+\sqrt{3})\text{ metres}
\]

4. Numerical value (optional):
\(\sqrt{3} \approx 1.732\), so
\[
h \approx 7(1 + 1.732) = 7 \times 2.732 \approx 19.1\text{ m}
\]

Answer: The height of the cable tower is \(7(1+\sqrt{3})\) metres (approximately \(19.1\) m).