Question

As seen from top of a 80m tall lighthouse, the angle of depression of two ships on the same side of light house in horizontal line with its base are 30° and 40° Find the distance between two ships.

Please answer fast⚡​

Answer 1

Correct question

As seen from top of a 80m tall lighthouse, the angle of depression of two ships on the same side of light house in horizontal line with its base are 30° and 60° Find the distance between two ships.

Answer

Let AB be the lighthouse and C and D be the position of two ships. Then,

AB = 80 cm,

∟ACB = 30°

∟ADB = 60°

Let CD be x metres,

Now, from ∆ DAB, we have

$\dashrightarrow \sf\dfrac{ AD}{AB} = cot 60\degree$

$\dashrightarrow \sf\dfrac{ AD}{80} = \dfrac{1}{ \sqrt{3} }$

$\dashrightarrow \sf AD= \dfrac{80}{ \sqrt{3} }$

Similarly,

From ∆ CAB, we have

$\dashrightarrow \sf\dfrac{CA}{AB} = cot 30\degree$

$\dashrightarrow \sf\dfrac{CD + DA}{AB} = cot 30\degree$

$\dashrightarrow \sf\dfrac{x+ \dfrac{80}{ \sqrt{3} } }{80} = \sqrt{3}$

$\dashrightarrow \sf x+ \dfrac{80}{ \sqrt{3} } = 80 \sqrt{3}$

$\dashrightarrow \sf \dfrac{ \sqrt{3} \: x+ 80}{ \sqrt{3} } = 80 \sqrt{3}$

$\dashrightarrow \sf \sqrt{3} \: x+ 80 = 80 \sqrt{3} \times \sqrt{3}$

$\dashrightarrow \sf \sqrt{3} \: x+ 80 = 240$

$\dashrightarrow \sf \sqrt{3} \: x = 240 - 80$

$\dashrightarrow \sf \sqrt{3} \: x = 160$

$\dashrightarrow \sf x = \dfrac{160}{ \sqrt{3}}$

Now, rationalise the denominator,

$\dashrightarrow \sf x = \dfrac{160 }{ \sqrt{3} } \times \dfrac{ \sqrt{3} }{ \sqrt{3} }$

$\dashrightarrow\sf x = \dfrac{160 \sqrt{3} }{3}$

$\dashrightarrow \underline{\boxed{ \sf x = 80 \sqrt{3} }}$

Hence, the distance between the ships is 80√3 metres.

Answer 2

Answer:

see the attachment above. hope it helps you..