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'.

Q: 37. 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'.

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

Some Applications of Trigonometry

Heights and distances - length of wire (30°)

Official Answer

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

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

Marking Scheme

11 mark: correct set-up $\cos 30° = \frac{6}{OB}$ (or equivalent using the right triangle) leading to $OB = \frac{6}{\cos 30°}$ .
2Award full credit for the final answer $4\sqrt{3}\text{ m}$ or its decimal equivalent $\approx 6.93\text{ m}$ ; accept $\frac{12}{\sqrt{3}}\text{ m}$ if clearly evaluated.

Hint

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

Quick Oral Answer

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

Analysis & 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.
$\cos 30° = \frac{OP}{OB}$ gives $OB = \frac{6}{\cos 30°} = \frac{12}{\sqrt{3}}$ , which rationalises to $4\sqrt{3}\text{ m} \approx 6.93\text{ m}$ .

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.

Common Mistakes

1Using $\tan 30°$ to compute the height of B and reporting that height ( $2\sqrt{3}\text{ m}$ ) as the wire length instead of the hypotenuse.
2Leaving the answer as $\frac{12}{\sqrt{3}}$ without rationalising to $4\sqrt{3}\text{ m}$ .
3Mixing up adjacent and opposite sides, e.g., writing $\sin 30° = \frac{6}{OB}$ and getting 12 m.

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.

$\cos 30° = \frac{\sqrt{3}}{2}$ is one of the standard-angle values students must memorise, and it produces the clean surd answer $4\sqrt{3}$ .

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.

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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 $\cos 30° = \frac{6}{\text{wire}}$ gives the wire length directly.

What is the exact value of the wire length?

It is $\frac{6}{\cos 30°} = \frac{12}{\sqrt{3}} = 4\sqrt{3}\text{ m}$ , 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.