As observed from the top of a light house, 100 m high above sea level, the angles of depression of ship, sailing directly towards it, changes from 30% to 60°. Find the distance travelled by the ship during the period of observation. (Use root 3 = 1.73)
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Answer:
In △ ABC ,
[tex] \frac{AB}{BC} = \tan60 \degree \\ \\ \implies \frac{100}{y} = \sqrt{3} \\ \\ \implies \: y = \frac{100}{ \sqrt{3} } \: \: \: ....(i)[/tex]
[tex]{\huge{\underline{\small{\mathbb{\pink{REFER \ TO \ THE \ ATTACHMENT}}}}}}[/tex]
In △ ABD,
[tex] \frac{AB}{BD} = \tan30 \degree \\ \\ \implies \: \frac{100}{y + x} = \frac{1}{ \sqrt{3} } \\ \\ \implies \: x + y = 100 \sqrt{3} \\ \\ \implies \: x = 100 \sqrt{3} - y \\ \\ \implies \: x = 100 \sqrt{3} - \frac{100}{ \sqrt{3} } \\ \\ = \frac{300 - 100}{ \sqrt{3} } = \frac{200}{ \sqrt{3} } \\ \\ (\sf \: from \: eq(i)) \\ \\ \implies \: x = \frac{200 \sqrt{3} }{3} = \frac{200 \times 1.73}{3} \\ \\ \boxed{ = 115.33m}[/tex]
the distance travelled by the ship during the period of observation is 115.33m
[tex] \rule{200pt}{2.5pt}[/tex]