Question 502 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.

Answer keywords:angle of elevationtangent ratiodistance ABsection of towerright triangleheights and distancesdifference of heights

Marking scheme

  1. 1.1 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.
  2. 2.1 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}.

Common mistakes

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

Analysis and 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.

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.

Question details

Q50 of 56 questions

Board
CBSE
Class
Class 10
Subject
Mathematics (Standard)
Year
2026
Section
Section E
Marks
2
Chapter
Some Applications of Trigonometry
Topic
Heights and Distances — Case Study (Radio Tower/Support-Wire Problem)
Bloom's level
apply
Syllabus topic
Unit 5: Trigonometry — Some Applications of Trigonometry: finding the height of an upper section of a tower using two angles of elevation from the same point
Author
CBSE
Reviewed by
Padhantu Academic Team

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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}.