Q50
2 marksSection E

Find the distance AB.

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 Q50
Some Applications of Trigonometry
Heights and Distances — Case Study (Radio Tower/Support-Wire Problem)
Official Answer

AB=PA−PB=63−23=43 m≈6.93 mAB = PA - PB = 6\sqrt{3} - 2\sqrt{3} = 4\sqrt{3}\text{ m} \approx 6.93\text{ m}, found using tan 60° and tan 30° with OP = 6 m and subtracting the lower section's height from the total height.

angle of elevationtangent ratiodistance ABsection of towerright triangleheights and distancesdifference of heights

Marking Scheme

  • 11 mark: correctly finding PB=23 mPB = 2\sqrt{3}\text{ m} and PA=63 mPA = 6\sqrt{3}\text{ m} using tan⁡θ=height/OP\tan \theta = \text{height}/OP for both angles of elevation.
  • 21 mark: correctly subtracting to get AB=PA−PB=43 mAB = PA - PB = 4\sqrt{3}\text{ m} (≈6.93 m\approx 6.93\text{ m}) with correct unit.

Hint

First find PB and PA separately using tan⁡(angle)=height/OP\tan(\text{angle}) = \text{height}/OP, then AB = PA − PB.

Quick Oral Answer

I find the heights of B and A separately using tan⁡(angle)=height/OP\tan(\text{angle}) = \text{height}/OP for each angle of elevation, then subtract PB from PA to get the length of the upper section ABAB, which is 43 m4\sqrt{3}\text{ m}.

Analysis & Explanation

Find AB, the length of only the upper section of the tower (from B to apex A).


Concept

  • OP = 6 m; angle of elevation of B = 30°; angle of elevation of A (apex) = 60°.
  • tan θ = height/OP gives the tower heights PB and PA in right triangles OPB and OPA.

Key points

  • AB is the difference PA − PB, not either height alone, since B lies between P and A.

Common mistakes

  • Reporting PA (total height) as AB.
  • Forgetting to rationalise 6/√3 to 2√3.

Common Mistakes

  1. 1Reporting PA (the full tower height) as the answer for AB instead of subtracting PB.
  2. 2Mixing up which angle of elevation (30° or 60°) corresponds to which point (B or A).
  3. 3Arithmetic slip while simplifying 6×(1/3)6 \times (1/\sqrt{3}) to 232\sqrt{3} (rationalising the surd).

Interesting Facts

Two-angle-of-elevation problems like this mirror how surveyors historically measured inaccessible heights using a theodolite from a single fixed ground station.

The 30°–60°–90°30°\text{–}60°\text{–}90° triangle used here has side ratios 1:3:21 : \sqrt{3} : 2, one of the two 'standard' triangles (along with 45°–45°–90°45°\text{–}45°\text{–}90°) that CBSE trigonometry problems are built around.

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Frequently Asked Questions

Why can't AB just be equal to PA?

PA is the total height from the ground to the apex of the tower, while AB is only the length of the upper section from point B to the apex. AB=PA−PB=43 mAB = PA - PB = 4\sqrt{3}\text{ m}.

What is the numeric value of AB?

AB=43 mAB = 4\sqrt{3}\text{ m}, which is approximately 6.93 m6.93\text{ m}.