Question

As observed from the top of a 75 m high light house from the sea level the angle 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 two ships.

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Answer 1

⠀⠀⠀⠀$\huge\underline{ \mathrm{ \purple{QUESTION}}}$

As observed from the top of a 75 m high light house from the sea level the angle 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 two ships

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⠀⠀⠀⠀⠀⠀$\huge{ \underline{ \blue{ \bold{ \underline{ \mathbf{AnSweR }}}}}}$

$\huge\underline{ \underline{ \green{ \mathrm{Given}}}} = >$

• Height of light house AD=75m

• And angle of depression of 1st ship <PAC=30°

• And angle of depression of 2nd ship <PAB= 30°

$\huge\underline{ \underline{ \green{ \mathrm{To\:Find}}}} = >$

We need to find BC the distance between the two ships.

$\huge\underline{ \underline{ \green{ \mathrm{Solution}}}} = >$

In right angled triangle ACD.

$\bold{ \bf{tan \: c = \frac{side \: opposite \: to \: < c}{side \: adjacent \: to \: < c} }} \\ \bold{ \bf{tan \: c = \frac{AD</p><p>}{</p><p>CD} }}$

⠀⠀⠀$=>\bold{ \bf{1 = \frac{75}{CD} }} \\ \\ => \bold { \bf{CD = 75m}}$

similarly,

In right angled triangle ABD

⠀⠀⠀⠀$=> \bf \: tan \: B = \frac{AD}{BD}$

⠀⠀⠀$=>\bf \frac{1}{ \sqrt{3} } = \frac{75}{BD} \\ \\ =>\bf BD= 75 \sqrt{3}$

⠀⠀⠀⠀$=>\bf \: BC + CD= 75 \sqrt{3} \\ \\ =>\bf \: BC + 75 = 75 \sqrt{3} \\ \\ => \bf \: BC= 75 \sqrt{3} - 75$

⠀⠀⠀⠀$=>\bf \: BC= 75( \sqrt{3} - 1)m$

Hence the distance between the two ships

=⠀⠀⠀⠀$\boxed{ \fbox{ \bold{ \bf{ \pink{ </strong><strong>BC</strong><strong>= 75 (\sqrt{3} - 1)} }}}}$

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hops this may help you

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$\huge \blue{ \mathfrak{thanks♡</p><p>}}$

Answer 2

Answer:

75✓3-1.....

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