Question 461 markSection E

  1. Case Study - 2. 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°. (i) Find the length of the wire from the point 'O' to the top of section 'B'.

Some Applications of Trigonometry
Heights and distances - length of wire (30°)

Official answer

OB=43 m≈6.93 mOB = 4\sqrt{3}\text{ m} \approx 6.93\text{ m}. Since OP = 6 m is adjacent to the 30° angle and OB is the hypotenuse, cos⁡30°=OPOB\cos 30° = \frac{OP}{OB} gives OB=6cos⁡30°=123=43 mOB = \frac{6}{\cos 30°} = \frac{12}{\sqrt{3}} = 4\sqrt{3}\text{ m}.

Answer keywords:4√3 m6.93 mcos 30°wire lengthhypotenuseOB = 6/cos30°12/√3angle of elevation 30°

Marking scheme

  1. 1.1 mark: correct set-up cos⁡30°=6OB\cos 30° = \frac{6}{OB} (or equivalent using the right triangle) leading to OB=6cos⁡30°OB = \frac{6}{\cos 30°}.
  2. 2.Award full credit for the final answer 43 m4\sqrt{3}\text{ m} or its decimal equivalent ≈6.93 m\approx 6.93\text{ m}; accept 123 m\frac{12}{\sqrt{3}}\text{ m} if clearly evaluated.

Hint

The 6 m ground distance is adjacent to the 30° angle and the wire is the hypotenuse, so use cos⁡30°=6÷(wire)\cos 30° = 6 \div (\text{wire}).

Quick oral answer

The wire is the hypotenuse and the 6 m ground distance is adjacent to the 30° angle, so OB=6cos⁡30°=123=43 m≈6.93 mOB = \frac{6}{\cos 30°} = \frac{12}{\sqrt{3}} = 4\sqrt{3}\text{ m} \approx 6.93\text{ m}.

Common mistakes

  1. 1.Using tan⁡30°\tan 30° to compute the height of B and reporting that height (23 m2\sqrt{3}\text{ m}) as the wire length instead of the hypotenuse.
  2. 2.Leaving the answer as 123\frac{12}{\sqrt{3}} without rationalising to 43 m4\sqrt{3}\text{ m}.
  3. 3.Mixing up adjacent and opposite sides, e.g., writing sin⁡30°=6OB\sin 30° = \frac{6}{OB} and getting 12 m.

Analysis and explanation

This part tests the direct use of the cosine ratio in the right triangle formed by the supporting wire, the tower, and the ground.


Concept

  • O is 6 m from the base; the wire OB is the hypotenuse, and the 30° angle of elevation is at O, so the 6 m ground distance is adjacent to it.
  • cos⁡30°=OPOB\cos 30° = \frac{OP}{OB} gives OB=6cos⁡30°=123OB = \frac{6}{\cos 30°} = \frac{12}{\sqrt{3}}, which rationalises to 43 m≈6.93 m4\sqrt{3}\text{ m} \approx 6.93\text{ m}.

Key points

  • Always rationalise the surd form (12/√3 → 4√3 m) before giving the final answer.

Common mistakes (परीक्षा में सावधानी)

  • Using tan 30° to find the tower's height and mistakenly reporting that as the wire length, or mismatching sine with the wrong side.

Real-world

  • Guy wires on real radio and cell towers are cut to length using exactly this cosine relation between the anchor point and the tower.

Interesting facts

Real broadcast masts are often 'guyed' towers held up by such angled steel wires (guy wires); the cosine rule used here is exactly how their lengths are specified.

cos⁡30°=32\cos 30° = \frac{\sqrt{3}}{2} is one of the standard-angle values students must memorise, and it produces the clean surd answer 434\sqrt{3}.

A 30° elevation is a shallow angle, so the wire (6.93 m) is only slightly longer than the 6 m ground run — a good sanity check on the answer.

Question details

Q46 of 53 questions

Board
CBSE
Class
Class 10
Subject
Mathematics (Basic)
Year
2026
Section
Section E
Marks
1
Chapter
Some Applications of Trigonometry
Topic
Heights and distances - length of wire (30°)
Bloom's level
apply
Syllabus topic
Unit 5: Trigonometry — Some Applications of Trigonometry: heights and distances using cosine of angle of elevation (CBSE Class 10 Mathematics)
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why do we use cosine here instead of tangent?

The wire is the hypotenuse and the known 6 m distance is the side adjacent to the 30° angle. Cosine relates adjacent and hypotenuse, so cos⁡30°=6wire\cos 30° = \frac{6}{\text{wire}} gives the wire length directly.

What is the exact value of the wire length?

It is 6cos⁡30°=123=43 m\frac{6}{\cos 30°} = \frac{12}{\sqrt{3}} = 4\sqrt{3}\text{ m}, which is approximately 6.93 m.

Is the wire longer or shorter than the 6 m ground distance?

Slightly longer — 6.93 m — because the wire is the sloping hypotenuse, always longer than the horizontal base.