Radio towers are used for transmitting a range of communication services including radio and television. The tower will either act as an antenna itself or support one or more antennas on its structure. On a similar concept, a radio station tower was built in two Sections A and B. Tower is supported by wires from a point O.
Distance between the base of the tower and point O is 36 cm. From point O, the angle of elevation of the top of the Section B is 30° and the angle of elevation of the top of Section A is 45°.
Generated by claude-sonnet-4-6 · 2026-06-15 10:37 · grounding stimulus
Model Answer
Let the base of the tower be P and PO = 36 cm.
(i) Length of wire from O to top of Section B:
Let top of Section B = Q. Angle of elevation = 30°.
$$\cos 30° = \frac{PO}{OQ} \Rightarrow \frac{\sqrt{3}}{2} = \frac{36}{OQ} \Rightarrow OQ = \frac{72}{\sqrt{3}} = 24\sqrt{3} \text{ cm}$$
Length of wire = 24√3 cm
(ii) Distance AB:
Height of Section B (PQ):
$$\tan 30° = \frac{PQ}{36} \Rightarrow PQ = \frac{36}{\sqrt{3}} = 12\sqrt{3} \text{ cm}$$
Height of top of Section A (PA):
$$\tan 45° = \frac{PA}{36} \Rightarrow PA = 36 \text{ cm}$$
$$AB = PA - PQ = 36 - 12\sqrt{3} = 12(3 - \sqrt{3}) \text{ cm}$$
(iii) Height of Section A from base:
$$PA = 36 \text{ cm}$$
Height of Section A = 36 cm
Source: Applications of Trigonometry, Chapter 9
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Explanation
- Part (i): Use cos ratio (hypotenuse = wire from O to top of B). cos 30° = base/hypotenuse.
- Part (ii): Use tan for both heights, then subtract PQ from PA to get AB. The key step examiners check is the subtraction and simplification 12(3−√3).
- Part (iii): tan 45° = 1, so height = horizontal distance directly. Simple but must be stated clearly.
- Always define your variables and show the trig ratio used — that's where method marks lie.