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ICSE Class X Math 2016 Paper

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Question : 29 of 57
Marks: +1, -0
An aeroplane at an altitude of 1500 metres finds that two ships are sailing towards it in the same direction. The angles of depression as observed from the aeroplane are 45∘45^{\circ} and 30∘30^{\circ} respectively. Find the distance between the two ships.
Solution:  
Let D is the aeroplane. A and B are two ships.
    Given:  CD=1500 m,∠EDA=30∘,∠EDB\;\;\text{Given:}\; CD=1500\,\text{m}, \angle EDA=30^{\circ}, \angle EDB   =45∘\;=45^{\circ}
In △DBC\triangle DBC ,
  tan⁡45∘=  DCBC=  1500BC\;\tan 45^{\circ}=\;\frac{DC}{BC}=\;\frac{1500}{BC}
  1=  1500BC\;1=\;\frac{1500}{BC}
  ⇒    BC=1500 m\;\Rightarrow\;\; BC=1500\,\text{m}
In △ADC\triangle ADC ,
tan⁡30∘  =  DCAC\tan 30^{\circ}\;=\;\frac{DC}{AC}
  13  =  1500AB+BC\;\frac{1}{\sqrt{3}}\;=\;\frac{1500}{AB+BC}
AB+BC  =15003AB+BC\;=1500\sqrt{3}
AB  =15003−1500AB\;=1500\sqrt{3}-1500
AB  =1500(3−1)AB\;=1500(\sqrt{3}-1)
  =1500(1⋅73−1)\;=1500(1\cdot73-1)
  =1500×0.73\;=1500 \times 0.73
  =1095 m.\;=1095\,\text{m}.
Hence, distance between two ships =1095 m=1095\,\text{m} .
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