A person in the helicopter flying at the height of 541 \(m\) above the river, observes two objects lying opposite to each other on either bank of a river. The angles of depression of the objects are 30\(^{\circ}\) and 45\(^{\circ}\). Then, find the width of the river.(Note: Substitute \(\sqrt{3} = 1.732\) if necessary.)
Answer:
Let \(A\) and \(B\) be the position of two objects lying opposite to each other on the bank of the river.
Let \(C\) be the position of the person flying in a helicopter.
In the right \(\triangle CDA\), \(tan \ \theta =\)
\(AD=\)\(m\).
\(DB=\)\(m\).
The width of the river is \(m\).