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ICSE Class X Math 2015 Paper

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Question : 40 of 52
Marks: +1, -0
The horizontal distance between two towers is 120 m120\text{ m}. The angle of elevation of the top and angle of depression of the bottom of the first tower as observed from the second tower is 30∘30^{\circ} and 24∘24^{\circ} respectively.
Find the height of the two towers. Give your answer correct to 3 significant figures.
Solution:  
Let ABA B and CDC D are two towers.
BD=120 m=EC,∠ACE=30∘,∠CBD=24∘  ". "  BD=120\text{ m}=EC, \angle ACE=30^{\circ}, \angle CBD=24^{\circ}\; \text{". "}\;
In right angle △CBD\triangle CBD ,
tan⁡24∘  =  CDBD  =  CD120\tan 24^{\circ}\;=\;\frac{CD}{BD}\;=\;\frac{CD}{120}
CD  =120×0.4452CD\;=120 \times 0.4452
  =53.42 m.\;=53.42\text{ m}.
In right angle △ACE\triangle ACE ,
tan⁡30∘  =  AEEC  =  AE120\tan 30^{\circ}\;=\;\frac{AE}{EC}\;=\;\frac{AE}{120}
AE  =120×0.5773AE\;=120 \times 0.5773
  =69.28 m\;=69.28\text{ m}
AB  =AE+EBAB\;= AE+ EB
  =AE+CD    [EB=CD]\;= AE+ CD \;\; [ EB= CD]
  =69.28+53.42\;=69.28+53.42
  =122.70 m.\;=122.70\text{ m}.
Height of 1st tower is 122.70 m122.70\text{ m} and 2nd tower is 53.42 m53.42\text{ m}.
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