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

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

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

Some Applications of Trigonometry

Applications of Trigonometry - Heights and Distances (Radio Tower)

Official Answer

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

wire lengthhypotenusecosine ratioangle of elevation 30 degreesradio towerright triangle

Marking Scheme

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

Hint

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

Quick Oral Answer

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

Analysis & 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

$\cos 30^\circ = PO/OB \Rightarrow OB = 6/\cos 30^\circ = 4\sqrt{3}\text{ m} \approx 6.93\text{ m}$ .
Cross-check via Pythagoras: $OB^2 = PO^2 + PB^2 = 36 + 12 = 48 \Rightarrow OB = \sqrt{48} = 4\sqrt{3}\text{ m}$ .

Common Mistakes

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

Common Mistakes

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

Interesting Facts

The value $4\sqrt{3}\text{ m}$ ( $\sim 6.93\text{ m}$ ) is a slightly longer wire than the vertical height of the lower section ( $2\sqrt{3}\text{ m} \approx 3.46\text{ m}$ ), 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.

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Frequently Asked Questions

Why use cosine and not tangent here?

Because the wire $OB$ is the hypotenuse of the right triangle, and cosine relates the adjacent side ( $PO$ ) to the hypotenuse ( $OB$ ), whereas tangent would only give the vertical height $PB$ .

Can Pythagoras theorem be used instead?

Yes — first find $PB = PO \tan 30^\circ = 2\sqrt{3}\text{ m}$ , then $OB = \sqrt{PO^2 + PB^2} = \sqrt{36+12} = \sqrt{48} = 4\sqrt{3}\text{ m}$ , giving the same answer.