Question

Q36 A moving boat is observed from the top of a 150m high cliff moving away from the cliff The angle
of depression of the boat changes from 60° to 45° in 2 minutes. Find the speed of the boat in m/min
(use V3- 1.73)
10

Answer 1

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Answer :-

$\rm{Let \: The \: Speed \: of \: The \: Be \: 'x \: ' \: m/min}$

$\rm{Distance \: Covered \: in \: 2min \: = \: 2x }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm \red{ \therefore CD= 2x}$

$\rm{Let \: BC \: Be \: 'y \: '}$

$\rm \blue{In \: \triangle \: ABC ,}$

$\small \rm{</p><p>\dfrac{AB}{BC}=tan60 \degree}$

$\small \rm{: \implies \dfrac{150}{y}=√3}$

$\small \rm{: \implies y = \dfrac{150}{√3}}$

$\small \rm{: \implies y = 50√3 \: ..............\: (1)}$

$\rm \blue{In \: \triangle \: ABD ,}$

$\small \rm{\dfrac{AB}{BD}=tan45 \degree}$

$\small \rm{: \implies \dfrac{150}{y+2x}=1}$

$\small{ \rm{: \implies y+2x = 150 \: ..............\: (2)}}$

$\rm{Substituting \: the \: Value \: of 'y \: ' \: from \: 1 \: and \: 2 \: Equn } \:$

$\small \rm{: \implies 50√3 + 2x = 150}$

$\: \small \rm: \implies 2x = 150 - 50√3$

$\: \small \rm: \implies 2x = 50 ( 3 - √3 )$

$\: \small \rm: \implies x = 25 ( 3 - √3 )$

$✦ \:\rm\pink {Speed \: of \: The \: Boat} =\rm{ 25( 3 - √3 ) m/min } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \:\: \: \: \: \: \: \: \: \: \: \: \: { \frac{</p><p>= 25 ( 3 - √3 ) \times 60}</p><p></p><p>{1000}}$

$\rm{: \implies \dfrac {3}{2}( 3 - √3 ) km / hr}$

$\: \therefore \rm{ \underline{ \underline \green{ speed \: of \: the \: boat \: is \: \: \dfrac {3}{2}( 3 - √3 ) km / hr }}}$

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