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

Right-angled figure showing tower foot P, midpoint B (top of lower section B) and apex A (top of upper section A) on a v — Fig. for Q49

Some Applications of Trigonometry

Heights and Distances — Case Study (Radio Tower/Support-Wire Problem)

Official Answer

$OA = 12\text{ m}$ — obtained from $\cos 60^\circ = OP/OA$ with $OP = 6\text{ m}$ .

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Marking Scheme

11 mark: correctly applying $\cos 60^\circ = OP/OA$ (or an equivalent correct method, e.g. finding PA via tan 60° then OA via Pythagoras/sine) to get OA = 12 m with the unit (m).

Hint

OA is the hypotenuse of right triangle OPA; use $\cos 60^\circ = OP/OA$ (adjacent/hypotenuse), with OP = 6 m.

Quick Oral Answer

OA is the hypotenuse of right triangle OPA, so I use $\cos 60^\circ = OP/OA$ , giving $OA = 6 \div \cos 60^\circ = 12\text{ m}$ .

Analysis & Explanation

Part (ii) of the radio-tower case study — find the wire length OA using the 60° angle of elevation.

Concept

Tower has two sections: A (top) above B, standing on foot P; wire runs from ground point O to A.
Given: OP = 6 m, angle of elevation of A from O = 60°.
In right triangle OPA (right angle at P), OP is adjacent to the 60° angle and OA (the wire) is the hypotenuse.

Key points

Correct ratio: $\cos 60^\circ = OP/OA$ (adjacent/hypotenuse) since OA is the hypotenuse, not a leg.

Common mistakes

Using tan 60°, which relates the two legs PA and OP, not the wire OA.
Confusing OA (wire, hypotenuse) with PA (tower height, a leg).

Common Mistakes

1Using $\tan 60^\circ$ (which relates PA and OP, the two legs) instead of $\cos 60^\circ$ (which directly relates OP and the hypotenuse OA).
2Confusing which point (A, the apex) has the 60° angle versus point B (which has 30°), since A is higher up the tower yet is asked about second.
3Omitting the unit (metres) in the final answer.

Interesting Facts

Guy-wire (support wire) calculations like this are used in real life to determine the exact cable length needed for radio masts and transmission towers before installation.

The sine and cosine ratios were first tabulated systematically by the Indian mathematician Aryabhata (5th century CE), whose 'ardha-jya' (half-chord) is the direct ancestor of the modern trigonometric functions used in this problem.

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Frequently Asked Questions

Why is cosine used here and not tangent?

Tangent relates the two legs of the right triangle (PA and OP), while OA itself is the hypotenuse. Since OP (adjacent) and the angle are known and the hypotenuse OA is required, cosine (adjacent/hypotenuse) is the direct ratio to use.

Could the sine ratio also be used?

Yes, indirectly: first find PA using $\tan 60^\circ = PA/OP$ ( $PA = 6\sqrt{3}\text{ m}$ ), then use $\sin 60^\circ = PA/OA$ to get the same answer, OA = 12 m. The cosine method is faster since it needs only one step.