Question 481 markVery Short AnswerSection E

Find the length of the wire from the point 'O' to the top of section 'B'.

Some Applications of Trigonometry
Applications of Trigonometry - Heights and Distances (Radio Tower)

Official answer

43 m4\sqrt{3}\text{ m} (≈6.93 m\approx 6.93\text{ m}) — using cos⁡30∘=PO/OB\cos 30^\circ = PO/OB in right triangle OPB, where PO = 6 m is the base and OB is the wire (hypotenuse), so OB=6/cos⁡30∘=43 mOB = 6/\cos 30^\circ = 4\sqrt{3}\text{ m}.

Answer keywords:wire lengthhypotenusecosine ratioangle of elevation 30 degreesradio towerright triangle

Marking scheme

  1. 1.1 mark: correct identification of the right triangle and ratio (cos⁡30∘=PO/OB\cos 30^\circ = PO/OB or use of Pythagoras), leading to the correct final answer OB=43 mOB = 4\sqrt{3}\text{ m} (≈6.93 m\approx 6.93\text{ m}).

Hint

The wire is the hypotenuse of the right triangle — use cos⁡30∘=base/hypotenuse\cos 30^\circ = \text{base}/\text{hypotenuse}, or Pythagoras with PO and PB.

Quick oral answer

Using cos⁡30∘=PO/OB\cos 30^\circ = PO/OB with PO = 6 m, the wire length OB=6/cos⁡30∘=43 mOB = 6/\cos 30^\circ = 4\sqrt{3}\text{ m}, approximately 6.93 m.

Common mistakes

  1. 1.Using tan⁡30∘\tan 30^\circ instead of cos30°, which gives only the height PBPB (23 m2\sqrt{3}\text{ m}) instead of the wire length OB.
  2. 2.Incorrectly simplifying 12/312/\sqrt{3} without rationalising, leaving the answer in a non-standard form.
  3. 3.Mixing up which angle (30° or 60°) belongs to section B versus section A.

Analysis and explanation

Asks for the hypotenuse (wire) of right triangle OPB, testing correct ratio choice.


Concept

  • Since the wire OB is the hypotenuse and PO = 6 m is the adjacent side to the 30° angle, cosine (adjacent/hypotenuse) is the correct ratio — not tangent, which gives only the vertical height PB.

Key Points

  • cos⁡30∘=PO/OB⇒OB=6/cos⁡30∘=43 m≈6.93 m\cos 30^\circ = PO/OB \Rightarrow OB = 6/\cos 30^\circ = 4\sqrt{3}\text{ m} \approx 6.93\text{ m}.
  • Cross-check via Pythagoras: OB2=PO2+PB2=36+12=48⇒OB=48=43 mOB^2 = PO^2 + PB^2 = 36 + 12 = 48 \Rightarrow OB = \sqrt{48} = 4\sqrt{3}\text{ m}.

Common Mistakes

  • Using tan30° instead of cos30° and wrongly reporting the height PB (2√3 m) as the wire length.

Interesting facts

The value 43 m4\sqrt{3}\text{ m} (∼6.93 m\sim 6.93\text{ m}) is a slightly longer wire than the vertical height of the lower section (23 m≈3.46 m2\sqrt{3}\text{ m} \approx 3.46\text{ m}), illustrating that a slanted support wire is always longer than the vertical height it supports.

Guy wires on real telecom towers are typically angled between 30° and 60° to the ground for optimal structural stability, matching the angles used in this problem.

Question details

Q48 of 56 questions

Board
CBSE
Class
Class 10
Subject
Mathematics (Standard)
Year
2026
Section
Section E
Marks
1
Type
Very Short Answer
Chapter
Some Applications of Trigonometry
Topic
Applications of Trigonometry - Heights and Distances (Radio Tower)
Bloom's level
apply
Syllabus topic
Unit V: Trigonometry — Some Applications of Trigonometry: finding length of an inclined line using angle of elevation
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why use cosine and not tangent here?

Because the wire OBOB is the hypotenuse of the right triangle, and cosine relates the adjacent side (POPO) to the hypotenuse (OBOB), whereas tangent would only give the vertical height PBPB.

Can Pythagoras theorem be used instead?

Yes — first find PB=POtan⁡30∘=23 mPB = PO \tan 30^\circ = 2\sqrt{3}\text{ m}, then OB=PO2+PB2=36+12=48=43 mOB = \sqrt{PO^2 + PB^2} = \sqrt{36+12} = \sqrt{48} = 4\sqrt{3}\text{ m}, giving the same answer.