Q48
2 marksSection E

  1. (iii) (a) Find the distance AB.

Some Applications of Trigonometry
Heights and distances - height of tower section AB
Official Answer

AB=43 m≈6.93 mAB = 4\sqrt{3} \text{ m} \approx 6.93 \text{ m}. Heights above the ground: AP=6tan⁡60°=63 mAP = 6 \tan 60° = 6\sqrt{3} \text{ m} and BP=6tan⁡30°=23 mBP = 6 \tan 30° = 2\sqrt{3} \text{ m}; distance AB=AP−BP=63−23=43 mAB = AP - BP = 6\sqrt{3} - 2\sqrt{3} = 4\sqrt{3} \text{ m}.

4√3 m6.93 mtan 60° = √3tan 30° = 1/√36√3 m2√3 mAB = AP − BPdifference of heights

Marking Scheme

  • 11 mark: correct heights of the two tops using tangent — AP=6tan⁡60°=63 mAP = 6 \tan 60° = 6\sqrt{3} \text{ m} and BP=6tan⁡30°=23 mBP = 6 \tan 30° = 2\sqrt{3} \text{ m} (½ mark each).
  • 21 mark: correct difference AB=63−23=43 m≈6.93 mAB = 6\sqrt{3} - 2\sqrt{3} = 4\sqrt{3} \text{ m} \approx 6.93 \text{ m}.
  • 3Deduct if the student subtracts wire (hypotenuse) lengths from parts (i)/(ii) instead of the vertical heights.

Hint

Find each height with tangent: top of A = 6tan⁡60°=63 m6 \tan 60° = 6\sqrt{3} \text{ m} and top of B = 6tan⁡30°=23 m6 \tan 30° = 2\sqrt{3} \text{ m}, then AB=63−23AB = 6\sqrt{3} - 2\sqrt{3}.

Quick Oral Answer

The top of A is 6tan⁡60°=63 m6 \tan 60° = 6\sqrt{3} \text{ m} high and the top of B is 6tan⁡30°=23 m6 \tan 30° = 2\sqrt{3} \text{ m} high, so the section AB=63−23=43 m≈6.93 mAB = 6\sqrt{3} - 2\sqrt{3} = 4\sqrt{3} \text{ m} \approx 6.93 \text{ m}.

Analysis & Explanation

This is the synthesis part of the case study, requiring two separate right-triangle heights and their difference.


Concept

  • Both tops are measured from the same ground point O at 6 m, so the tangent ratio gives their vertical heights: AP=6tan⁡60°=63 mAP = 6 \tan 60° = 6\sqrt{3} \text{ m} and BP=6tan⁡30°=23 mBP = 6 \tan 30° = 2\sqrt{3} \text{ m}.
  • The vertical gap AB=AP−BP=63−23=43 m≈6.93 mAB = AP - BP = 6\sqrt{3} - 2\sqrt{3} = 4\sqrt{3} \text{ m} \approx 6.93 \text{ m}.

Key points

  • Both heights share the same base of 6 m, which is exactly what makes the subtraction valid.

Common mistakes (परीक्षा में सावधानी)

  • Subtracting the wire (hypotenuse) lengths from parts (i)/(ii) instead of the vertical heights — AB is a vertical segment, so only tan-based heights may be subtracted.

Real-world

  • This 'difference of two elevations' technique is the standard method for measuring the height of an upper storey, flagpole, or antenna without climbing it.

Common Mistakes

  1. 1Subtracting the wire lengths (12 − 6.93) from parts (i) and (ii) instead of the vertical heights, giving a wrong AB.
  2. 2Using cos or sin instead of tan to find the heights, since AB is a vertical distance found from opposite/adjacent.
  3. 3Forgetting to rationalise 63\frac{6}{\sqrt{3}} to 2√3, then mishandling the subtraction 63−236\sqrt{3} - 2\sqrt{3}.

Interesting Facts

The double-angle setup 30° and 60° is deliberately chosen so that tan⁡60°=3\tan 60° = \sqrt{3} and tan⁡30°=13\tan 30° = \frac{1}{\sqrt{3}} combine to give a clean 434\sqrt{3} answer.

Surveyors measure the height of an antenna or an upper floor by exactly this 'difference of two angles of elevation' method, avoiding any need to climb the structure.

Interestingly, AB (43 m4\sqrt{3} \text{ m}) equals the wire length OB from part (i) — a neat coincidence arising from the 30°–60° geometry.

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Frequently Asked Questions

Why do we use tangent and not cosine for AB?

AB is a vertical distance. Tangent relates the vertical height (opposite) to the horizontal base (adjacent), so tan gives each height, and their difference is AB.

Can we subtract the wire lengths from parts (i) and (ii)?

No. The wires are sloping hypotenuses, not vertical. AB is vertical, so we must subtract the vertical heights 63 m6\sqrt{3} \text{ m} and 23 m2\sqrt{3} \text{ m}.

What is the exact length of AB?

AB=63−23=43 mAB = 6\sqrt{3} - 2\sqrt{3} = 4\sqrt{3} \text{ m}, which is approximately 6.93 m.