Q1. [4] § 9.1 Heights and Distances § 9.2 Summary
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' (as shown in figure). Distance between the base of the tower and point 'O' is 6 m. 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 60°.
Based on the above information, answer the following questions :
- (i) Find the length of the wire from the point 'O' to the top of section 'B'. [1]
- (ii) Find the length of the wire from the point 'O' to the top of section 'A'. [1]
- (iii) Find the distance AB. [2]
Previously asked in CBSE board exam
2026 30/1/1 Q37
Generated by claude-sonnet-4-6 · 2026-06-15 10:36 · grounding stimulus
Model Answer
Given: PO = 6 m, angle of elevation of B = 30°, angle of elevation of A = 60°
(i) Length of wire OB:
In △OPB: cos 30° = PO/OB
$$OB = \frac{PO}{\cos 30°} = \frac{6}{\frac{\sqrt{3}}{2}} = \frac{12}{\sqrt{3}} = 4\sqrt{3} \text{ m}$$
(ii) Length of wire OA:
In △OPA: cos 60° = PO/OA
$$OA = \frac{PO}{\cos 60°} = \frac{6}{\frac{1}{2}} = 12 \text{ m}$$
(iii) Distance AB:
$$PB = PO \times \tan 30° = 6 \times \frac{1}{\sqrt{3}} = 2\sqrt{3} \text{ m}$$
$$PA = PO \times \tan 60° = 6 \times \sqrt{3} = 6\sqrt{3} \text{ m}$$
$$AB = PA - PB = 6\sqrt{3} - 2\sqrt{3} = 4\sqrt{3} \text{ m}$$
Source: Applications of Trigonometry, Chapter 9
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Explanation
- For wire length, use cos θ = base/hypotenuse (hypotenuse = wire = OB or OA).
- For heights PB and PA, use tan θ = height/base.
- AB = PA − PB since B is below A on the same vertical tower.
- Examiner awards 1 mark each for (i) and (ii), and in (iii): 1 mark for finding both heights correctly + 1 mark for the subtraction. Show all steps clearly.
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