Question

As observed from the top of a 100 m high light house from the sea level, the angles of depression of two ships are 30°and 45°.if one ship is exactly behind the other on the same side of the light house, find the distance between the two ships. (Use root3=1.732)
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Answer 1

This is ur answer buddy ❤️✌

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\therefore{\text{Distance\:between\:two\:trees=73.2 m}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{ \underline \bold{given : }} \\ : \implies \text{height \: of \: light\:house = 100 \: m} \\ \\ : \implies \text{angle \: of \: depression \: = 45 \degree \: and \: 30 \degree} \\ \\ \red{ \underline \bold{to \: find : }} \\ \implies \text{distance \: between \: two \: ships = ?}$

• According to given question :

$\text{In \: right }\triangle \text{ABC} \\ : \implies tan \theta = \frac{p}{b} \\ \\ : \implies tan \: 45 \degree = \frac{100}{b} \\ \\ : \implies 1 = \frac{100}{BC} \\ \\ : \implies \text{BC = 100 \: m} \\ \\ \text{In \: right }\triangle \text{ABD} \\ : \implies tan \: \theta = \frac{p}{b} \\ \\ : \implies tan \: 30 \degree = \frac{100}{bd} \\ \\ : \implies \frac{1}{ \sqrt{3} } = \frac{100}{BD} \\ \\ : \implies \text{BD = 100} \sqrt{3} \: m \\ \\ \bold{For \: distance \: between \: two\: ships}\\ : \implies Distance = BD- BC\\ \\ : \implies Distance =100 \sqrt{3}-100 \\ \\ : \implies Distance = 100 ( \sqrt{3}-1 ) \\ \\ : \implies Distance = 100 \times 0.732 \\ \\ \green{ : \implies \text{Distance}\approx 73.2 m}$