As observed from top of a 150 m tall light house- the angles of depression of two ships approaching it are 30- and 45- If one ship is directly behind the other- find the distance between the two ships-

IN ∆ ACD tan45° =AD/DC = gt; 1 = 150/DC = gt; DC = 150 m in ∆ ABD, tan30° = AD/DB = gt; 1/√(3) = 150/DB = gt; DB = 150√(3) = gt; DB = 150√(3) ...

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Standard X

Mathematics

Angle of Depression

As observed f...

Question

As observed from top of a 150 m tall light house, the angles of depression of two ships approaching it are $30^{\circ}$ and $45^{\circ}$ . If one ship is directly behind the other, find the distance between the two ships.

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Solution

IN ∆ ACD

tan45° = $\frac{A D}{D C}$

=> 1 = $\frac{150}{D C}$

=> DC = 150 m

in ∆ ABD,

tan30° = $\frac{A D}{D B}$

=> $\frac{1}{\sqrt{3}} = \frac{150}{D B}$

=> DB = 150 $\sqrt{3}$

=> DB = 150 $\sqrt{3}$ m

DB = DC + CD

=> 150 $\sqrt{3}$ = 150 + CD

=> (150 $\sqrt{3}$ - 150) = CD

=> 150(1.732 - 1) = CD
[ $\sqrt{3}$ = 1.732 ]

=> CD = 150×0.732

=> CD = 109.8 m

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As observed from top of a 150 m tall light house, the angles of depression of two ships approaching it are 30∘ and 45∘. If one ship is directly behind the other, find the distance between the two ships.

IN ∆ ACD tan45° =ADDC => 1 = 150DC => DC = 150 m in ∆ ABD, tan30° = ADDB => 1√3=150DB => DB = 150√3 => DB = 150√3 m DB = DC + CD => 150√3 = 150 + CD => (150√3 - 150) = CD => 150(1.732 - 1) = CD [ √3 = 1.732 ] => CD = 150×0.732 => CD = 109.8 m

As observed from top of a 150 m tall light house, the angles of depression of two ships approaching it are $30^{\circ}$ and $45^{\circ}$ . If one ship is directly behind the other, find the distance between the two ships.