Question

A boy standing on a horizontal plane finds a bird flying at a distance of 100 m from him at an elevation of 30°. A girl standing on the roof of a 20 m high building, finds the elevation of the same bird to be 45°. The boy and the girl are on the opposite side of the bird. Find the distance of the bird from the girl. (Given√2 = 1414)

Answer 1

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$\rm \: Let \: C \: be \: the \: position \: of \: Bird, \: and \: B \: and \: E \: are \: the \: position \: of \: boy \: and \: girl \: respectively.$

$\rm \: Let \: the \: height \: of \: building \: be \: DE \: = 20 \: m$

$\rm \: Let \: the \: distance \: of \: girl \: from \: bird \: be \: EC = x \: m$

$\rm \: In \: right \: \triangle \: BAC,$

$\rm \: sin \: 30 \degree = \frac{AC}{BC}$

$\longrightarrow \: \rm \: \frac{1}{2} = \frac{AC}{100}$

$\longrightarrow \: \rm \: AC = \frac{100}{2} = 50 \: m$

$\rm \: Now, \: CF = AC - AF$

$\rm = 50 - 20 = 30 \: m \: \: \: \: \: \: \: ( \because \: AF = DE)$

$\rm \: In \: right \: \triangle \: EFC,$

$\rm \: sin \: 45\degree = \frac{CF}{CE}$

$\longrightarrow \: \rm \frac{1}{ \sqrt{2} } = \frac{30}{x}$

$\longrightarrow \: \rm \: x = 30 \times \sqrt{2}$

$\rm \: \: \: \: \: \: \: \: \: \: \: \: \: = 30 \times 1.414$

$\rm \: \: \: \: \: \: \: \: \: \: \: \: \: = 42.42 \: m$

$\rm \: Hence, \: the \: distance \: of \: girl \: from \: bird \: is \: \red{42.42 \: m.}$