Q8 of 61 Page 215

A person in an helicopter flying at a height of 700 m, observes two objects lying opposite to each other on either bank of a river. The angles of depression of the objects are 30° and 45°. Find the width of the river. (√3 = 1.732 )


Given, AD = 700 m and BC = ?


In triangle ACD,


∠ACD = ∠ MAC (alternate angles are equal)


= 45°


We know,





⇒ DC = 700 m …… (1)


In triangle ABD,


∠ABD = ∠ OAB


= 30° (alternate angles are equal)


We know,





⇒ BD = 700√3 …… (2)


Now, adding equation (1) and (2)–


Width of the river = BD + DC


= 700 + 700√3


= 700(1 + √3)


= 700 (1 + 1.732)


= 700 × 2.732


= 1912.40 m


∴ Width of the river is 1912.40 m


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